- Drawing dot structures
- Drawing Lewis diagrams
- Worked example: Lewis diagram of formaldehyde (CH₂O)
- Worked example: Lewis diagram of the cyanide ion (CN⁻)
- Worked example: Lewis diagram of xenon difluoride (XeF₂)
- Exceptions to the octet rule
- Counting valence electrons
- Lewis diagrams
- Resonance and dot structures
- Formal charge
- Formal charge and dot structures
- Worked example: Using formal charges to evaluate nonequivalent resonance structures
- Resonance and formal charge
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- More on the dot structure for sulfur dioxide
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
- Molecular polarity
- 2015 AP Chemistry free response 2d and e
In this video, we apply VSEPR theory to molecules and ions with three groups or “clouds” of electrons around the central atom. To minimize repulsions, three electron clouds will always adopt a trigonal planar electron geometry. If none of the clouds is a lone pair, the molecular geometry will also be trigonal planar. If one of the clouds is a lone pair, the molecular geometry will be bent. Created by Jay.
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- At9:15, why don't both the oxygen's make double bonds with sulfur? I know this breaks the octet rule, but sulfur is in the 3rd period, so isn't it more important to have no formal charge than obey the octet rule?(95 votes)
- It is because Oxygen is more electronegative, and therefore will consume more of the electrons. Sulfur, being less electronegative, cannot have more electrons around it than the more electronegative oxygen(66 votes)
- Can anyone clarify the last rule of 'ignoring lone pairs of electrons'?
I thought that the lone pairs of electrons had a slightly stronger repulsive force than the bonded electrons and that they would reduce the bond angle between the 2 Oxygens to less than 120 (I know he denotes it as ~120 degrees). Surely in order to get the 'bent' shape, we have to specifically consider the lone electron pairs as opposed to ignoring them?(17 votes)
- The lone pairs help determine the electron pair geometry.
Thus, one lone pair and two bonding pairs give a trigonal planar electron geometry with bond angles of about 120°.
But molecular geometry considers only the bonding pairs.
Thus SO₂ has a trigonal planar electron geometry, but the two S-O bonds are at an angle of about 119°.
Since we consider only the bonds, the molecular geometry is bent.(25 votes)
- I don't get that if a central atom can complete an octet then why is the formal charge an issue?
I mean what happens due to the neutralization of the formal charge? How does it contribute to the stability? Or is formal charge a greater stability factor than octet?(20 votes)
- But like in the video it you have too much electrons in a molecule and you can't put them in any other thing that's bonded to the center, put some electrons in the center.(0 votes)
- Why O-S=O and not O=S=O?(15 votes)
- At12:04it says bent, so is it 120 degrees or 104.5 degrees?(4 votes)
- In this instance, the bond angle will be 120 degrees. If there was a lone pair on the central atom, then the bond angle would be reduced to 104.5 degrees. See the video titled "VSEPR for 4 electron clouds" for an example of this(4 votes)
- At11:10he says the bond angle for sulfur dioxide (bent) is 120 degrees, the same as for a trigonal planar structure. In my chemistry course, however, my teacher explained that the lone electron pair on the sulfur atom will exert a stronger repulsive force and thus push the oxygen atoms further away, resulting in a smaller bond angle of about 115 degrees.
Can someone please explain? Thanks in advance.(2 votes)
- Your teacher is partly correct. Lone pairs do take up more space than bonds (or "push" the bonds away from them, if you like to think of it that way). However, the bond angle is 119.3°, not 115°. But please be aware that when we have lone pairs "pushing" the bonds away from them, the number of degrees they shift by is not a set thing that would apply to all molecules -- in some molecules or ions the amount of "pushing" is quite small, in others it is more pronounced. It just depends on the details.
(Your teacher may have been thinking of the nitrite anion, NO₂¯, which does have a bond angle of about 115°)(4 votes)
- At3:00he mentions that Boron doesn't need to follow the octet rule but it can. Does that mean that BF3 would have four resonance structures? The one that doesn't follow the octet rule AND the three that do?(4 votes)
- It cannot because of the number of total Valence Electrons. Boron can follow the octet rule, but the total number of Valence Electrons prevents this. Boron normally "wants" 6-8 electrons.(0 votes)
- for evaluating the formal charge on oxygen u did it like 6-7 n got a -1 thn for sulfur y r we substracting 6 from 5? it shud have been 5-6 . as for oxygen u substracted its group no from the no of electrons it is surrouned by but the opposite was followed for sulfur.. i did nt get it(2 votes)
- in BF3.......doesnt boron lose 3 electrons which leaves 2 electrons tthus filling the first shell(1 vote)
- No, the electrons are shared. The compounds of boron are not often covered in General Chemistry because they often involve different types of bonds than those usually covered at this level of study.
Also note that boron does not obey the octet rule (at least not in the usual sense).(4 votes)
- Why aren't electrons on the terminal atoms considered in determining electron geometry?(1 vote)
- We can't "see" the electrons with our instruments, but we can see the atoms and measure their bond lengths and bond angles.
That’s why we ignore the lone pair electrons in determining molecular geometry.
We must, however, consider the lone pairs in determining the electron geometry about an atom.(4 votes)
If I want to draw a dot structure for boron trifluoride, I need to think about VSEPR theory-- so valence shell electron pair repulsion. The valence electrons are going to repel each other and force the molecule into a particular shape or geometry. And so we'll first start off by drawing the dot structure. So for boron trifluoride, I find boron on the periodic table. It's in group three. So it has three valence electrons. Fluorine is in group seven. So it has seven valence electrons. I have three of them. So 7 times 3 gives me 21. 21 plus 3 gives me 24 valence electrons that we need to account for in our dot structure. Boron is less electronegative than fluorine. So boron's going to go in the center like that. And the boron is bonded to three fluorine atoms. Let me go ahead and put in those three fluorine atoms like that. So we just represented six valence electrons. So here's 2, 4, and then 6. So 24 minus 6 gives me 18 valence electrons left that we need to worry about. We're going to put those leftover electrons on a terminal atoms, which are our fluorines in this case. Fluorine follows the octet rule. So each fluorine is now surrounded by two valence electrons. So each fluorine needs six more to have an octet of electrons. So I'll go ahead and put six more valence electrons on each of the three fluorines. And 6 times 3 is 18. So we just represented 18 more valence electrons. And so now we are all set. We've represented all 24 valence electrons in our dot structure. And some of you might think, well, boron is not following the octet rule here. And that is true. It's OK for boron not to follow the octet rule. And to think about why, let's assign a formal charge to our boron atom here. And so, remember, each covalent bond consists of two electrons. Let me go ahead and draw in those electrons in blue here. And when you're assigning formal charge, remember how to do that. You take the number of valence electrons in the free atoms, which is three. From that number you subtract the number of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that's OK. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when we talk about our next step here for predicting the shape. So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron clouds are on the same plane. So I'm going to draw sheet of paper here, a plane. And we're going to put our boron atom in the center. And here's one of our electron clouds. And then here's another one, and here's our third one here. So they're going to get as far away from each other as they possibly can. We call this shape trigonal planar. So let me go ahead and write that here. So this is a trigonal planar geometry of my electron clouds surrounding my central atom. And since we don't have any lone pairs of electrons to worry about on our central atom, we can go ahead and predict the geometry of the molecule as being the same here as the geometry of the electron cloud. So the molecule has a trigonal planar shape, as well. Now, when we think about bond angles, right? So the easiest way to think about what's the bond angle for trigonal planar or what would we predict the bond angle to be, you think about a circle. And since the electrons repel each other equally, you want to divide your circle into three equivalent angles. So 360 degrees divided by 3 is 120. So you could think about all these bond angles here as being 120 degrees. And so you could think about those electron clouds repelling each other equally. So we have a trigonal planar geometry with a bond angle of 120 degrees. So three electron clouds. Let's do another example of a molecule that has three electron clouds-- so in this case, sulfur dioxide. So sulfur dioxide, let's count up the number of valence electrons. Sulfur's in group six, so six valence electrons. Oxygen is also in group six. 6 times 2 is, of course, 12. 12 plus 6 is 18. So we have 18 valence electrons for our dot structure. Sulfur is less electronegative, right? If you look at a periodic table, so sulfur is below oxygen. So sulfur's going to go in the center. Sulfur is bonded to two oxygens like that. So that just represented four valence electrons, right? Here's two. And here's two more. So 18 minus 4 is 14 valence electrons left. All right, so we're going to start assigning some those leftover electrons to our terminal atoms, which are our oxygens. Oxygen's going to follow the octet rule. And so, therefore, each oxygen gets six more electrons to give each oxygen an octet. So we go ahead and do that. So I just represented 12 more valence electrons, right? Six on each oxygen. So 14 minus 12 is 2 valence electrons. So I have two valence electrons left over. And, remember, when you have leftover valence electrons, you go ahead and put them on your central atom. So we can go ahead and put those two valence electrons here on our central atom like that. Now, we're not quite done with our dot structure because sulfur doesn't have an octet. That's one way of thinking about it. You could also think about formal charge. Sulfur's formal charge is not minimized in this dot structure. So we need to share some electrons, right? So I could take, let's say-- let me go ahead and make these blue here. So I could take a lone pair of electrons from either oxygen. I'm just going to say that we take a lone pair of electrons from that oxygen and move them in here to form a double bond between the sulfur and the oxygen. So if I do that, now I have a double bond between the sulfur and the oxygen. The oxygen on the right now has only two lone pairs of electrons around it. The oxygen on the left still has three lone pairs of electrons around it like that. And the sulfur still has a lone pair of electrons here in the center. All right, so if we assign formal charges now-- let's go ahead and do that really fast. So we know that we have electrons in these bonds here. And so if we assign a formal charge to the oxygen on the left-- let's do that one first, all right, so this oxygen on the left here. All right, oxygen normally has six valence electrons in the free atom. And in our dot structure, we give one of these electrons in blue to the oxygen and one to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail about resonance structures and formal charge. We talked about those in earlier videos. So make sure that you watch them. But this is our final dot structure here. So let me go ahead and redraw it here. This is one of the possible dot structures you can draw sulfur dioxide that fulfills our rules for drawing dot structures. So let me go ahead and put in these valence electrons here so we can see it a little bit better. All right, so I'll leave out formal charges now because we're just focusing in on geometry. We're concerned about VSEPR theory. So we have our dot structure. We go back up to check our steps for predicting the shape. And now we are going to count the number of electron clouds that surround our central atoms, the regions of electron density, which could be valence electrons in bonds, so bonding electrons, or non-bonding electrons, like lone pairs of electrons. So if we look at our central atom, which is sulfur, and let's see if we can count up our electron clouds. So this is an electron cloud over here. There's a region of electron density. So we have one electron cloud. Over here on the right, this double bond, we can consider it as an electron cloud. We're not worried about numbers of electrons, just regions of them. And then for the first time, we now have a lone pair of electrons, right? And this, we can also think about as occupying an electron cloud. So we have three electron clouds. And we saw in the previous example that when you have three electron clouds, the electron clouds are going to try to adopt a trigonal planar shape. So I could redraw this dot structure and attempt to show it in more of a trigonal planar shape here. So let's go ahead and show it looking like this-- once again, not worried about drawing informal charges here-- so something like this for the structure. Let me go ahead and put those electrons in our orbital here so we can see that electron cloud a little bit better. And so, once again, our electron clouds are in a trigonal planar geometry. So we would expect those bond angles to be approximately 120 degrees, right? So let me go ahead and put this in here. So approximately 120 degrees, it's probably slightly less than that. But that is what we would predict the geometry of the electron cloud to be. So let's go back up and look at our rules here, our steps for predicting the shapes of molecules. So we've done step three, right? We have predicted the geometry of the electron clouds around our central atom. And now we go on to step four here. We're going to ignore any lone pairs on our central atom when we predict the geometry of the overall molecule. And so that now pertains to our example here. We're going to ignore that lone pair of electrons on the sulfur. And we're going to ignore this lone pair of electrons when we're talking about the shape of the molecule. So even though the electron clouds have a trigonal planar geometry, we say that the shape of the molecule has a bent or angular shape. And so if you look at that, if you just look at the atoms, if you ignore the lone pairs and look at the atoms, you'll see this kind of bent or angular shape here. And so that's what we say is the shape of the molecule. So you could say bent, or you could say angular here. All right, so that's two examples of molecules with three electron clouds. And remember, it's not just the number of electron clouds. You have to ignore lone pairs of electrons to predict the final shape of the molecule.