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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond .
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then would all the orbitals become sp^3 hybrids given the steric number would now be 4 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why does jay draw lines connecting both sets of the p orbital lobes when only one set of lobes has electrons ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side .
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does hybridisation occur before or during bonding ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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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 three atoms . we only need three of our orbitals .
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also , do all atoms undergo hybridisation if they are bonded to another atom ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create .
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why are the three pairs of electrons per f atoms in bf3 not counted as lone pairs ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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which steric number corresponds to what type of hybridization form ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals .
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why does the remaining p orbital electron do a pi bond and not a sigma bond ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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each one of those sp2 hybrid orbitals has one electron in it . 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 .
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why does carbon atom have 2 purple orbitals when it has only one unhybridised electron left ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon .
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why do n't all the 2p orbitals take part in hybridization ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
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is it necessary that an hybridized orbital is necessary for pi bonds ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example .
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please explain to me why we are taking the sp3 hybridized state of c when getting to sp2 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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is the steric number only used in finding the # of hybridized orbitals for the central atom of the molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero .
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are there 0 lone pairs referring to the 0 lone pairs surrounding the boron atom specifically ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why do we ignore the s orbital while drawing the shape of molecules ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character .
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does the fluorine hybridize to bond with boron ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons .
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like it said in overall , is it you always fill valence electrons into sp2 orbitals first then free p orbital right ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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around is steric number only accounted by number of the sigma bonds of one carbon atom or total number of sigma bonds in the whole molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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here we discuss the hybridization of central atoms , are the more-exterior atoms ( i 'm considering the fluorine atoms in bf3 ) similarly hybridized ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why does boron have an empty p orbital ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in .
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in the beginning , why is there only one electron in the 2s orbital when it should be filled first ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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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 valance electrons .
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but the sp3 hybrid orbital has 75 % p character and the sp2 hybrid orbital has 67 % p character , then how does the sp2 hybrid orbital have a higher p character ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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can you plz tell me how can we easily learn what is hybridization ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that .
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why does jay say that there is only 1 pi bond ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
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are pi bonds made up of more than 2 valence electrons ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in .
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but what is the motive of orbitals getting hybridized ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here .
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does hybridisation actually lends the atom more stability ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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does hybridization actually take place or it 's just a hypothetical concept to explain extra valencies and the like ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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why does a double bond indicate sp2 hybridization ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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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 three atoms . we only need three of our orbitals .
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even though carbon is only bonded to three other atoms , are n't there still four bonds being formed ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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would it not be more energy efficient to just demote all three of the p orbitals to the sp orbital and leave the s orbital alone ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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would the easiest way of finding the amount of p in your hybrids be to look at how many bonds the central atom is in ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe .
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why is the lower lobe of the sp2 orbitals ignored in the videos ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now .
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earlier it was said to avoid confusion but is it because it has lower electron density than the upper lobe ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell .
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why do you only need 2 p orbitals ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now .
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can sp hybridisation and the drawing of organic isomers be used used in conjunction with each other so as to show how the electron clouds of each atom ( connected to a central atom ) affect the structure of the molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero .
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how doesbf3 not have any lone pairs of electrons ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept .
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can steric number be used to find the hybridization of any molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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can it be used to determine hybridization of ions ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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if a compound accepts an electron to become an ion , will its hybridization change ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react .
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are all sp2 atoms from group 3 on the periodic table capable of accepting an electron pair as mentioned ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why the unhybridized orbital is the p orbital instead of s ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals .
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fluor has lone electrons of 3 orbitals for each pair , so why it is not showed here , like for ammonium and water in previous video ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero .
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is bf3 considered an sp2 hybredization because of the lone pairs and their ability to create double bonds through resonance ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here .
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in c2h4 molecule , why do we need hybridisation in the first place ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule .
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how do the angles of atoms and molecules interact with the reactions , structure , and general stats ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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how co2 has sp2 hybridization while it has linear geometery ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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when we took the 2s orbital and promoted it and then we left a p orbital unhybridized why did n't we take the 3 valence electrons the 3 in the p orbital and left the unexcited s unhybridized ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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so does that mean that the number of orbitals that we should take for hybridisation should be equal to number of atoms the carbon or whichever element there is , is being bonded with ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create .
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didnt boron have 2s2 and 2p1 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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does hybridization occur in all molecules ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here .
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can a molecule be non-hybrid or whatever ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create .
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why does boron not need an octet ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why is there an unbonded p orbital in the ethylene molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell .
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so u should there be two unhybridised p orbitals in sp hybrisisation ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why should the s orbital be promoted ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell .
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why cant three p orbitals be considered ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital .
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does the hybrid orbital mean that if there is more than 10 is it still a sp10 orbital or does it not go that high ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms .
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if the distance between the carbon atoms in ethylene is shorter than those in ethane due to increased s character , as stated then are the sigma bonds with hydrogen atoms going to be shorter than those in ethane ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
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can a hybridised orbital form a pi bond ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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how only one p orbital remain unhybridized of carbon in a ethylene molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three .
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if the n=3 shell can hold max 18 electrons then why does calcium have 8 electrons in its 3rd shell and 2 electrons in its 4th shell ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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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 valance electrons .
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how is the sp2 hybridation supposed to work ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell .
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in the case of c2h4 , why not hybridize all the p orbitals to get 4 sp3 orbitals ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals .
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how are we supposed to know which of these situations is most energetically favorable ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing .
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how we gon na determine steric number ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why is there one electron in the 2s orbital and three electrons in the p orbital ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here .
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after he makes the lewis dot structure for bf3 , why is there no double bond with one of the fluorine atoms to create an octet around the boron atom ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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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 three atoms . we only need three of our orbitals .
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all the atoms must fill an octet to be stable , correct ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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with the bf3 why is the electrons around the f not counted as single electron pairs ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why have we left one p orbital unhybridised ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why does n't the carbon atom just use its 3 electrons in the p orbital ( instead of using 1 from 2s and 2 from 3p , while keeping another p hanging around ) ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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if the steric number is 3 , is it always 3 sp^2 hybrid orbitals and 1 non-hybrid p orbital ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization .
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in c2h4 why cant just carbon undergo sp3 hybridisation and and each carbon atom form 2 sigma bond with 2 hydrogen atoms and 2 sigma bond with the other carbon atom ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now .
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what 's the vsepr theory and when you 're drawing your molecular structure does it have to be in a specific pattern , like ammonia ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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how come sn is 3 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
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in c2h4 structure they are 5 sigma bonds and 1 pi bonds but in structure 2 pi bonds are laterally overlapped on other then why we are counting as 1 pi bond ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
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how could electrons make pi bonds even they have the same charge ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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why do you only count the number of sigma bonds on one atom when referring to steric number ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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what is hybrid state of fluorine in bf3 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon .
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why does the 2s orbitals are not full ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
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in general , does hybridization always occur ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in .
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why in c2h4 all the orbitals are hybridized because in h2o and nh3 we see that all orbitals are hybridized ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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hi , what happens to the angle of bond in bf3 when boron accepts electrons and forms a lone pair ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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for consistency , like the previous example of c2h4 which gave the percentage of s hybridization and p hypridization , what is the % of hybridization for bf3 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 .
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does the hybridization have a 67 % mix of s characteristic and 33 % p characteristic ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals .
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how do i know all the distances between the elements , as in the minute 0 ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
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i thought a pi bond created a double bond , but there is also a sigma bond , so why is there not three bonds ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
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so the amount of atoms the carbon is attatched to is the number of orbitals that are needed ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron .
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what is the use of finding the steric number ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here .
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is so2 a bent molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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where did bf3 get the unhybridized p orbital from ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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so bf3 is actually more ( re ) active because of the missing electrons in the empty p orbital showing potential , yes or no ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
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why is the # of lone pairs on bf3 zero ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell .
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why ca n't the two carbon atoms share two sp3 hydride orbitals and then share the two other with hydrogen atoms ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond .
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i still do n't understand how do we know how to know which atom is in the front and which one is in the back when drawings the shape of any molecule ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization .
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how come the argument `` carbon bonding with three atoms '' for having 3 hybrid orbitals is valid ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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we have carbons four , valence electron represented . one , two , three and four . 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 three atoms .
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why ca n't the double bond be made of two sp3 hybrid orbitals ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
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why are there 2 unhybridized p orbitals drawn in ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in .
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around i was wondering , what if we were to switch the the carbon into another element like sulfur , would we fill up the hybridized orbitals with 6 arrows , and two of them being down ?
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: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . 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 three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . 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 . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . 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 . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here 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 once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then 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 . 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 . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have 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 rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and 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 . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching 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 . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . 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 valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
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when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
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i was taught that a double bond is a pi bond ( and only a pi bond ) not a sigma bond + a pi bond , which is correct ?
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