hi there I'm Jeremy Krug and we are beginning unit 2 Section 7 which is about how we visualize chemical bonds and visualize these molecules in three dimensions let's Jump Right In here and talk about what chemical bonds can look like so let's imagine that we have a single Bond so imagine we have two carbon atoms like this and there's a single Bond normally what you would see would be that the two electrons are being shared pretty much straight in between those two atoms right there which is probably what you would have expected now let's take a look at a double bond if we have a carbon-carbon a double bond once you notice that this is a little bit different we have two bonds that have to be placed in there somehow and that first bond is going to be shared right in the middle just like it was in the single but your second Bond here is going to be shared in a somewhat different way we have a some this this sort of a looped configuration and so that is what your double bond is for the most part going to look like now what about a triple bond because somehow in a triple bond you have to be able to fit in not one or two but actually three bonds into that little space there well here is how that would work you'd have one of the bonds that's that's in that straight conformation just like we had in the others but now we have a looped Bond going in this direction and then there's a second looped Bond that's actually and this is kind of hard to tell it's actually coming out at us it is in that third dimension and so we have two looped bonds and one straight bond in a triple bond now the straight bonds are given a name those are called Sigma bonds and we represent that with the Greek letter Sigma this little lowercase Sigma that kind of looks like a cursive O and notice that every one of these bonds has exactly one Sigma Bond now the double bond and the triple bond have these looped bonds those are called Pi bonds just like the Greek letter Pi that you use in math so you have that lowercase pi and a double bond has one of these Pi bonds a triple bond has a sigma and two Pi bonds so that's how these bonds are basically configured Sigma bonds just so you know are called that because they're the result of overlapping s orbitals as we've learned previously s orbitals are spherical in shape whereas Pi bonds are the result of overlapping P orbitals and so they have that more of a looped shape uh and so we have Sigma and Pi now some observations about the bonds I tried to draw it in this way but just so that everyone is on the same page about this notice that I drew the single bonds as the longest of the bonds triple bonds are the shortest Bond length double bonds are in the middle if you look at the way I had them configured I tried to draw them like that but make sure that you are aware of that single bonds are the longest triple bonds are the shortest every single bond is a sigma Bond every double bond has one Sigma and one Pi Bond and every triple bond has one Sigma Bond and two Pi bonds now with that information right there you should be able to count up the number of Sigma and Pi Bonds in any molecule that anyone could draw for you now we're going to try a couple examples of this let's take a look at hcn and we're going to figure out how many Sigma bonds and Pi bonds there are here well hopefully you see that there's a single Bond so that's a sigma and then there's a triple bond which is one Sigma and 2 pi so when you add it up it's two Sigma and two Pi bonds so hopefully that's not too difficult what if I were to give you a much more complex molecule something like this maybe some sort of a an organic molecule well it's pretty much the same thing you notice that there are a lot of single Bonds in here in fact every one of those single bonds is going to be a sigma Bond all basically all six of those single Bonds are sigmas there are a couple of double Bonds in here that I see as well and each of those double bonds this one as well as that one will have one Sigma and one Pi so we have those in there and then there's a triple bond in here which is a sigma and 2 pi so when you add up these sigmas and pies hopefully you can see that we have a total of nine Sigma bonds and four Pi bonds so that's how you can count up Sigma bonds and Pi bonds now another very important part of being able to uh to visualize molecules is to understand that there's something called hybridization of orbitals in these molecules now if we think about a fairly simple molecule like a methane one that we've probably drawn a couple times already it has this structure we have the carbon atom in the middle the central atom and then we have four Sigma bonds four single bonds there and we have that molecule but if we start thinking about this we might think that there's something wrong with this structure now what could be wrong with this structure well let's think about the electron configuration of carbon carbon is there in the middle it has four valence electrons that's the electron configuration of carbon and the valence electrons are the only ones that are participating in the bonding so the ones we care about are the 2s2 and the 2p2 electrons so when we plot those on an orbital diagram notation here we have 2s which has the two electrons there and 2p following hunz rule is going to look like this and based upon what I see here I only see two electrons that are looking for a partner it looks like there are only two electrons that are available for bonding and so if carbon can only make two bonds how is it possible that carbon is drawn here making four bonds how is that possible well it makes us Wonder does this molecule exist well it certainly does exist and this is the correct Lewis electron dot diagram so what's going on well the carbon atom in the center there is undergoing something called hybridization its orbitals are hybridized now here's how this works we have these these four orbitals that we're looking at here there's one s orbital and we have three p orbitals and they go through a hybridization process and then we have these four electrons and they're you know each each of those electrons is able to occupy a hybridized orbital so now we have four electrons in four hybridized orbitals now what are these orbitals well they're not fully s and they're not fully P either in fact we can say that they are one part s and three parts P or as we might say s p 3 hybridized and so we would say that the carbon atom in the methane molecule has sp3 hybridization now how can we determine that for any Central atom in a molecule because pretty much any molecule that has a central atom is going to do something like this well here's how you do it it's actually very simple if you've been following along with Sigma bonds and Pi bonds and understanding how these structures work all you have to do is add the number of Sigma bonds on an atom to the number of unshared electron pairs that are touching that atom and when you do that you're going to get a number that's in between 2 and 6. now that number actually has a name that's called the steric number and we sometimes use that to talk about electron geometry as we're going to here in our next video but if that number ends up being 2 then we say the hybridization is sp one part S one part P if the sum ends up being 3 the hybridization is SP2 if the sum is 4 then the hybridization is sp3 if it happens to be 5 then we go up to sp3 D hybridization and if the sum happens to be six we call it sp3 D2 hybridization now we're going to do a few examples with this let's see if we can determine the hybridization of the central atom in each of these molecules so here's hcn we've looked at this molecule a few times already so once again focus on the central atom how many Sigma bonds do we have here well there's one and then there's one over here that's 2. How many unshared pairs on that carbon well I don't see any so 2 plus 0 is 2. so the stearic number is two so what's the hybridization s p s p hybridization that's all you have to do now how about water well how many Sigma bonds are on that Central atom right there well I see two there's one and then there's two so we have two and how many unshared pairs well we have two there's one right there and one right there so two plus two equals four so what's the hybridization if the steric number is four that chart tells us it is sp3 and how about this structure here we have the nitrate polyatomic ion so what's the hybridization of that nitrogen in the middle well how many Sigma bonds one two three because one of those is a pi Bond and doesn't count any unshared pairs on the nitrogen there aren't are there so three plus zero equals three and that's direct number of three gets us a hybridization of SP2 so that's how we do hybridization it's not hard and just so you know on the AP exam if you're taking AP Chemistry they're not going to to use any hybridizations Beyond sp3 on the exam so if you learn something about Sigma and Pi bonds or something about hybridization please shoot me a thumbs up and I hope to see you in the next video where we're going to continue with unit 2 Section 7 learning about molecular geometry and bond angles