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AP.Chem:

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now that we understand hybridization states let's do a couple of examples all right and so we're going to identify the hybridization States and predict the geometries for all the atoms in this molecule except for hydrogen and so let's start with this carbon right here so the fast way of identifying a hybridization state is to say ok that carbon has a double bond to it therefore it must be sp2 hybridized and if it's sp2 hybridized we know the geometry around that carbon must be trigonal planar with bond angles approximately 120 degrees this carbon over here right is also I also has a double bond to it so it's also sp2 hybridized with trigonal planar geometry all right let's move to this carbon right here so that carbon has only single bonds around it and the fast way of doing it is it is if you see all single bonds it must be sp3 hybridized and if that carbon is sp3 hybridized we know the geometry is tetrahedral all right so tetrahedral geometry with an ideal bond angles of 109.5 degrees around that carbon alright let's move over to this carbon right here so this carbon has a triple bond on the right side of it and so the fast way of doing of doing this is if it has a triple bond it must be SP hybridized here so SP hybridized and therefore the geometry would be linear right with a bond angle of 180 degrees and same with this carbon this carbon has a triple bond to it so it also must be SP hybridized with linear geometry and so that's why I drew it this way so it's linear around those two carbons here let's go ahead and count up count up the total number of Sigma and PI bonds for this so that's also something we talked about in the previous videos here so first let's count out the number of Sigma bonds alright so let's go back over to to start with this carbon here so here's a sigma bond to that carbon here's a sigma bond to that carbon we know that a double bond one of those bonds is a sigma bond and one of those bonds is a PI bonds let me go ahead and also draw on our PI bonds in red so already already coloured the Sigma bond blue so let's say this one is the PI bond alright let's continue assigning assigning all of our bonds here so I know this single bond is a sigma bond I know this single bond is a sigma bond right so all of these single bonds here are Sigma when I get to the triple bond I know one of those is a sigma bond and two of those are PI bonds right so two of those are PI bonds here and then finally I have one more bond it's a single bond so I know that it is a sigma bond here and if you count up all of those all of those Sigma bonds right you should get 10 so let's do that really quickly so you get let me go ahead and change colors here so you get one two three four five six seven eight nine and ten so we have a ten Sigma bonds total and in terms of pi bonds we had three PI bonds so three PI bonds for this molecule you can also find hybridization States using a steric number so let's go ahead and do that really quickly so let's go back to this carbon and let's find the hybridization state of that carbon using steric number so let's use let's use green for this so steric number is equal to the number of Sigma bonds plus lone pairs of electrons alright so one two three sigma bonds around that carbon so three plus zero gives me a steric number of three therefore I need three hybrid orbitals and sp2 hybridization gives me three hybrid orbitals alright if I want to do for this carbon I would have one two three four alright so the steric number would be equal to four sigma bonds and zero lone pairs of electrons giving me a total of four from my steric number so I need four hybrid orbitals I have four sp3 hybridized orbitals at that carbon and then finally let's do it for let's do it for this carbon right here so using steric number steric number is equal to number of Sigma bonds plus numbers of lone pairs of electrons so there are two sigma bonds around that carbon zero lone pairs of electrons steric number of two means I need to hybrid orbitals and an SP hybridisation that's what you get you get two SP hybridized orbitals like that alright let's move on to another example let's let's do a similar analysis and before we do I notice I excluded hydrogen here and that's because hydrogen is only bonded to one other atom so there's no real geometry to talk about all right let's move on to this example so this molecule is diethyl ether and let's start with let's start with this carbon right here so the hybridization state all right well the fast way of doing it is to notice that there are only single bonds around that carbon right only Sigma bonds and so therefore we know that carbon is sp3 hybridized with tetrahedral geometry so sp3 hybridized tetrahedral geometry alright let's look at this carbon right here it's the exact same situation right only Sigma or single bonds around it so this carbon is also sp3 hybridized right and so therefore tetrahedral geometry let's next look at the oxygen here so if I wanted to figure out the hybridization and the geometry of this oxygen steric number is useful here so let's go ahead and calculate the steric number of this oxygen so that's number of Sigma bonds so here's a single bond so that's a sigma bond and then here's another one so i have two sigma bonds so two plus number of lone pairs of electrons around the atom so here's a lone pair of electrons and here's a lone pair of electrons so i have two lone pairs of electrons so two plus two gives me a steric number of four so i need four hybridized orbitals for this oxygen and we know that occurs when you have sp3 hybridization so therefore this oxygen is sp3 hybridized or four sp3 hybrid orbitals around that oxygen all right let's do geometry of this oxygen so the the electron groups right there are four electron groups around that oxygen so each electron group is in an sp3 hybridized orbital the geometry of those electron groups might be tetrahedral but not the geometry around that the oxygen here so the geometry around the oxygen and if you ignore the lone pairs of electrons right you can see that it is bent all right so even though that oxygen is sp3 hybridized its geometry is not tetrahedral the geometry of that oxygen there is bent or angular alright and because of symmetry alright this carbon right here is the same as is the same as this carbon so it's also sp3 hybridized and then this carbon over here is the same as this carbon so it's also sp3 hybridized so symmetry made our lives easy on this one alright let's do one more example so once again our goal is to find the hybridization States and the geometries for all the atoms except for hydrogen and so once again let's start with carbon let's start with this carbon right here alright so once again our goal is to find the hybridization state so the fast way of doing it is to notice that there's a there's one double bond to that carbon so it must be sp2 hybridized and therefore the geometry is trigonal planar so trigonal planar geometry alright let's do let's do the steric number way so if I were to calculate the steric number right steric number is equal to number of Sigma bonds so here's a sigma bond here's a sigma bond I have a double bond between the carbon and the oxygen so one of those is a sigma bond right and one of those is a pi bond which I'll draw and read here so I have three sigma bonds around that carbon so three plus zero lone pairs of electrons gives me a steric number of three so I need three hybridized orbitals and so once again sp2 hybridization alright let's do let's do the next carbon so let's move on to this one alright so I see only single bonds around that carbon therefore it must be sp3 hybridized with tetrahedral geometry so sp3 hybridized tetrahedral geometry same thing for this carbon right all only single bonds around it only Sigma bonds so it's sp3 hybridized with tetrahedral geometry let's finally look at this nitrogen here alright so if I want to find the hybridization state of this nitrogen I could use steric number so the steric number is equal to number of Sigma bonds so around this nitrogen here's a sigma bond it's a single bond right here's another one and here's another one so I have 3 Sigma bonds I have one lone pair of electrons alright so three plus one gives me four steric number four means I need four hybridized orbitals and that's our situation with sp3 hybridization and so this nitrogen is sp3 hybridized but it's tetra heat it's it's geometry is not tetrahedral alright so the geometry for that nitrogen as we discuss in earlier video right so it has these three sigma bonds like this and a lone pair of electrons and that lone pair of electrons is at an sp3 hybridized orbital and if we look at that geometry right and ignore the lone pair of electrons because you always ignore the lone pairs of electrons when you're looking at geometry we can see we have this sort of shape here so the nitrogen is bonded to three atoms the carbon hydrogen and hydrogen and then we have this sort of a shape like that alright so in the back there and you can see we call this trigonal pyramidal all right so the geometry around that nitrogen is trigonal pyramidal all right so that does it for three examples of organic hybridization so practice a lot for this

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