How to Determine Molecular Geometry
PROFESSOR: In this video, I want to cover how to determine the 3D shapes of a molecule. So the three shapes of molecule can take many forms. It depends on two factors. So there's two factors that go into determining the shape of a molecule-- 3D shapes of a molecule.
And so the only two factors are, number one, the number of lone pairs of electrons-- number of lone pairs of electrons. And the other factor is the number of atoms surrounding the center atom-- surrounding the center atom. So in this case, if we had H2O, we had the molecule, that we would represent that as H, O, H.
So in this case, if we were referring to the number of atoms surrounding the center atom, we were referring to this atom and this atom. So in that case, we would have two. So the easiest and most effective way to determine the shapes of molecules, I find, is simply to do a table.
And so I've included one here. I can show you my method. OK, so as you can see here, this diagram represents-- the x's are atoms, and these are lone pairs of electrons. These represent lone pairs.
So if we're given any molecule, the easiest way to determine whether it's tetrahedral or trigonal planer or octahedral is to go like this. So because we have those two factors, you'll want to just do a simple diagram. We'll have number of lone pairs.
And on this side, we will have the number-- I'll call them valance atoms even though that's probably not the correct term. Number of valance atoms-- so like in the H2O, the two H's would be the valance atoms. So in this case, we'll just make a quick table.
So the number of lone pairs, as you can see, we can have zero lone pairs. But you know, we do need at least two valance atoms because if we just had H2, we would just look like this-- H, H.
Obviously, this doesn't have any kind of shape. It's just a line. So no matter how we orient that in 3D space or rectilinear space or however you want to say it, it's just going to be a line. So if we have two, then we have the possibility of it being bent as in this case. So that means we need to start with two valence atoms. [INAUDIBLE] I'll just move this.
I'll put this in a different color. So I'll have two. And so we can go all the way up to three, four, five. So I'll go up to six in this case. And the number of lone pairs, we can start at 0.
We can have one loan pair. We can have two lone pairs, or we can have three. So there is one other thing that we need to define. We have to define the prefixes for the numbers-- so the number prefixes.
Now we should know-- and this is very important. And we have to know that-- I'm going to put this in a different color. So we need to know that bi equals two and bri equals three. Tetra equals four, and even though octa-- we'll write octa.
Even though that represents eight, for these purposes and only in these purposes will we refer to this as six. And you'll see why in a few minutes, and you'll kind of understand the intuition of why I'm referring to octa more earnestly as six even though it's supposed to be 8. So as you can see here, if we take any number of valence atoms-- so let's say we have-- well, let's say we have H2O.
Then that would be two valence atoms. Oops-- two valence atoms. And we would have one lone pair, right, for H2O. So we would have to-- let's see, one. So if we look over in this chart, we can see that it would be in one plane. But it would be bent. So we're going to go-- because these are trigonal planar.
Anything here trigonal planar is anything over here. Tetrahedral-- all of these. So like, if this was tetrahedral, this is tetrahedral, this is tetrahedral. These are kind of like sub shapes sort of. And so if this is bi-- or trigonal, bipyramidal, then this is-- all of these will be that as well. Octahedral-- these are all octahedral.
Anyway, so as you can see, if we have two valance atoms and we have one lone pair, then this would be trigonal planar. So I'll just write trigon for short. I'll write trigonal planar.
So now let's say we had three valance atoms. And again, let's say we have zero lone pairs. Well, then we look over here. And we say, OK, well, here's three valence atoms and zero lone pairs.
Oh, that's trigonal planar as well. So this is trigonal. And I actually am going to start. Well, I'll write planer. So as you can see, these two are-- no, let's do one more example.
Let's say we have-- let's say we have two valence atoms and three lone-- or actually let's say we have two valence atoms and two lone pairs-- two valence atoms, two lone pairs. We look over here, and we see that this one has two valence atoms and two lone pairs.
So this is tetrahedral. Now you might be wondering, well, how do you come up with this on your own? How do you have a system of this when you don't have a reference chart?
Well, it's easy because we've defined the prefixes. This one has two. This one is three, you know, four, and six. We can say, OK, look at this.
If we have to and we add one, we got trigonal planer. Well, what's 2 plus 1? Well, obviously 2 plus 1 equals 3.
Well, what do we have for three? Well that equals trigonal or tri. So we can assume that if anything equals 3, we'll get trigonal planer.
So what if we have three and zero? Well 3 plus 0 is 3. And if we look over here, tri equals 3. So we're going to have a trigonal planer.
All you have to do is count the number of valence atoms and the number of lone pairs. So you just add lone pairs and add the number of valance atoms. I'm going to write that so you can read it-- number of valance atoms.
And that gives you when you match-- and then match the prefix. And so just to confirm, like, what is this, I mean does this work for everything? I mean, we haven't defined what five is.
So what if we have-- let's say we have three valence atoms and we have two lone pairs. Well, we can't manage that with a prefix. But what happens-- OK, so this is three.
We're taking three atoms plus two pairs. That equals 5. But if we sub in a prefix for this, if we go tri plus bi for two-- we're just subbing in the prefixes right here-- then we're going to get tri plus bi.
Well, maybe that has something to do with trigonal bipyramidal, middle bipyramidal. I'm not actually sure how to pronounce that properly. But basically, this would be-- we're just adding these prefixes. So this is going to be trigonal bi-- no, that's an I. My mistake-- bipyramidal.
So we can say that, OK, so 3 plus 2-- OK, so it's going to be-- let's just double check this. Three valance atoms-- as you can see, we have three valence atoms and two lone pairs. Total equals 5.
Then this is going to be-- I'm just going to do tri bi for short. Then will that work for, say, three lone pairs and two atoms? Well let's see.
We have three lone pairs. We're going to have two valance atoms. Then we look over here. And look, we have three valence atoms-- or, sorry, three lone pairs, two valence atoms. And that is going to be tri bi as well. So this will work.
I mean, if we have-- like for anything, like, as you can see, it's either trigonal or things like that. Like, if it has two-- like, we had two here and we had two lone pairs and two valence electrons, that equals tetrahedral because 2 plus 2 equals 4, and our prefix is four. So what about octa?
So if we have, say, five valence atoms and one lone pair, will we still get the same results? Well, if we have 5 here, we have 1, 2, 3, 4, 5. And then we have one lone pair. Total is six.
Well, you can say it. We can just write this as octa. Will this work for the opposite?
I mean, what if we had two loan pairs and, say, four valence atoms. Well, look at what we have here. Two lone pairs-- and we have 1, 2, 3, 4 valence items.
So two and four works as well. We could just write octa. And so you can see all you have to do is add them, find out what the prefix is, knowing that even though is six instead of eight. But you just add these, the pairs and the valence atoms.
I mean, you can do a Lewis structure first to find out how many lone pairs there are. For instance, like with H2O, we can see that-- because there's a lone pair-- that we're going to have a bent molecule. But if we just want to know whether it's trigonal planar, all we have to do is add the lone pairs, add the valence electrons.
Now we also have to look at these subshapes because these are kind of subshapes. I mean, it could be trigonal planar. But is it a bent trigonal planer, or is it even or? There's like 120 degree angles everywhere.
Is there tetrahedrals with, you know, a loan pair or not? And you can-- I mean, really, once you know whether it's tetrahedral and trigonal pyramidal, then-- I think I did pronounce that right, pyramidal-- then you can just draw a structure. For instance, octahedral here-- if we look at this one here, then to draw that in real life, we're taking one atom.
We have a center atom. We're going down. And this is going to be behind.
It's going to come in front. And it's going to be like an x. I, mean this is not a very good three dimensional drawing, but imagine this is being in front and these are being behind the plane of the page.
Then you can kind of draw that and say, well, these are going to be in a plane, and these are going to be sticking out the top. And if we add a square planar, if we look at this one here, then that's going to be in a square. And-- oops, I'm going to draw this a bit better. And then the loan electron pairs are going to come out the top.
But the atoms here are going to be within the same plane. So you can kind of very easily determine whether it's trigonal planar or octahedral. But then you have to use common sense to figure out, well, these ones are going to be in the same plane. They're going to be square planar.
But they are going to be octahedral as well. But the subshape is going to be square planer. And so, I mean, the basic rules apply.
When we're dealing with electrons and the electrons are trying to repel, so we're just trying to find out what the most common sense situation for these atoms to be in because they're all trying to repel. Like, this atom here is trying to repel this atom, and this atom is trying to repel that. But then they're also trying to repel the lone pairs.
And so as you can see, this is the most stable formation where it's the least stress on the electrons here. They're as far away as possible from each other as you can go. So anyway, this is basically the simplest way, I think, to find out whether or not it's linear or trigonal planar.
You don't have to memorize the configurations. Just know that if they add up to four, it's going to be tetrahedral. If they add up to six, it's going to be octahedral.
And if it goes up to bi, of course it's going to be linear because it only has two unless it's bent with an electron pair. But that would mean that it would have three. So then it's tri. It's not bi. Anyway, I hope this makes it a bit more simple.