sp² hybridization | AP Chemistry | Khan Academy
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sp² hybridization | AP Chemistry | Khan Academy

Voiceover: 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 don’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 gonna 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 gonna promote the S orbital up and this time, we only
need two of the P orbitals. We’re gonna take one of the P’s and then another one of the P’s here. That is gonna 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 gonna 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
gonna have an effect on the length of the bonds that we’re gonna 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 gonna 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 don’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 gonna 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 lone 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 gonna 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 wanna 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 gonna 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 doesn’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.

About James Carlton

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22 thoughts on “sp² hybridization | AP Chemistry | Khan Academy

  1. wait a second… the electron configuration of carbon is 2s2, 2p2… why do you have it as 2p3? 0:50 . this breaks Hunds rule, doesnt it? can someone explain.?

  2. I'd have a question about NO3- in which N is sp2 hybridized. How many electrons are there on the s orbital? Are there 2 electrons on the s, and 3 lonely electrons on the p orbitals or can one from the 's' jump onto another higher orbital? Or how does this hybridization happen in case of the nitrate ion? Can it be that the s (with 2 e-) and the first two p (with 1 e- + 1e-) orbitals combine , these will be the hybrid orbitals which make the sigma bonds* and the one left p orbital makes the 'pi' bond..? (*the s electrons are given to one of the oxygen atoms..) Or is it totally wrong? Could someone explain how it is? Thanks in advance!

  3. How did he know to combine the hydrogen s orbital to the sp2 hybrid orbital of the carbon though, and not the p orbital (When he was drawing the structure for ethene)? Cause like, what's wrong with overlapping the hydrogen s orbital with the "spare/unfilled" carbon p orbital?

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