Rob Lederer: So here is some stuff that we know. CH4, there is a formula for methane gas and there is a Lewis diagram for it right there. We know that if we build it, we are going to have that's four effected pairs around the central carbon, and each one is a bond to a hydrogen, we've got a tetrahedral shape. Okay. Now this tetrahedral shape - if you are in tune here to what we've been talking about. Now you are thinking, well you know what? Since carbon is 1S2 2S2 2P2, not really though, what carbon likes to do in its box diagram here, it likes to kick an electron from the 2S2 to the 2P2, so we're got a one, two, three, four lone electrons available in these orbitals. Hey, guess what we do. We bond a hydrogen to each one of these orbitals to get CH4. Four valance orbitals here; one, two, three, four. There they go. The hydrogen donates its electron and shares it in between and S orbital and three P orbital, you know what that means? That means now one of this is an overlap of hydrogen to an S orbital and 2 S orbital or carbon. Then the other ones are 2 P orbitals, but that wouldn't mean that three of the bonds are alike and one of them is different. It has a different nature or character because of the different shape of the S orbital to the Ps.
No, what really happens is that all of these bonds are entirely equivalent. If they are entirely equivalent, that means then that there isn't an S in three Ps. Well, it does, but here is what happens. We take the one S orbital and the three P orbitals and we stick them together all along and we come up with one, two, three, four, equivalent orbitals when they bond. So we say that those four valance orbitals come together, 1 S and 3 Ps and hybridize into one distinct type of orbital, so four effective pairs around the central atom means that, that central atom has SP3 hybridization. So CH4 has SP3 hybridization. NH3 would be this with a lone pair on top. So it's got four effective pairs. When you see four effective pairs, that means that the bones are all equivalent in the valance orbital that this is an equivalent to these orbitals here, that's called SP3 hybridization around the central atom.
Now here is -- okay, that's C2H4. Remember that you count multiple bonds as one, so this is one, two, three effective pairs. When you have three effective pairs, the shape around that central atom is trigonal planer. Well, this is also a central atom, so it's also that trigonal planer shape. Trigonal planer and three effective pairs around the central atom is going to be that you have SP2 hybridization. You know what that means? One S and two of the P orbitals get together all along and they form three equivalent orbitals. What that means then that there is one other P orbital that doesn't hybridize. Here is what it all looks like. Here is CH4 and here is the Lewis diagram. Here is a diagram that has one, two, three, hybridized orbitals; one S and two Ps hybridized together to make a 120 degree separation and that's why we're getting the trigonal planer shape.
So this carbon has it, and this one has it too. The bond to one of the one S orbitals of hydrogen here with a hybridized orbital here and a bond to the other carbon with another hybridized orbital, but you've got a double bond. Where do you get the double bond from? This whole green here, this and this together is the double lower bridge, means one orbital, that is the unhybridized orbital for this carbon and it is unhybridized here. They have one electron in them each, just like all these orbitals all around here have. This forms another bond and makes up one bond, this makes up the other, and you get the double bond. One of the bonds in a double bond is always a hybridized one and one of them is always an unhybridized. This is called because it is the unhybridized orbital, it's called a pi bond and this one because it's a hybridized orbital, it's called a sigma bond. So you've got sigma and pi.
Now here a triple bond. That's coming. A triple bond gives you a linear shape but you've only got two effective pairs. So when you've got a carbon bond to a carbon with a triple bond, hydrogens in other end, that's ethane. Then you are going to have one, two, three, well, that's one effective pair and 1S2; that gives you a linear structure, and two effective pairs is SP hybridization. Okay. Now what is that? That is a pipeline here and here, they are from unhybridized orbitals. Remember, there is four in the orbitals. This one S and one P here, they hybridize and that is one of our hybridized orbitals. Here for carbon and other one here and one in the middle to bond to the other carbon, but there is two unhybridized P orbitals. One forms a bond and one forms another bond to get the triple bond. So there is your hybridization for simple ones, and then come five effective and six effective pairs.
When you've got five effective pairs around the central atom, like phosphorous has here, what kind of hybridization this, that have for all of these bonds? Well, what, if you got five, you've got, well, an S orbital and three P orbitals all coming together in hybridize, but that's only four orbitals. You need five here. So what actually happens remember we can exceed the architect rule because of the presence of D orbitals. So one D orbital has to come into the mix in order to then amalgamy with the three S, the one S orbital -- I mean the three P orbitals form five hybridized orbitals. So one of the Ds and one S but three of the Ds, and this is called DSP3 hybridization. So I have to remember, when you have five effective pairs, the hybridization is DSP3; one, two, three, four, five, we should be able to count up, right. For six effective pairs, you're going to have an S orbital, three Ps and two Ds, and so along. They amalgamy to form one type of hybrid orbital. Two D orbitals and S and 3 Ps, D2SP3, two, three, four, five, six hybridized odorless, and that is the various types of hybridization that you can have; hybridization.
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