- Dot Structures Questions
- Drawing dot structures
- Formal charge and dot structures
- Resonance and dot structures
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
VSEPR for 4 electron clouds
In this video, we apply VSEPR theory to molecules and ions with four groups or “clouds” of electrons around the central atom. To minimize repulsions, four electron clouds will always adopt a tetrahedral electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be tetrahedral (no lone pairs), trigonal pyramidal (one lone pair), or bent (two lone pairs). Created by Jay.
Want to join the conversation?
- at6:07it is easy to understand how the unpaired electrons repel the hydrogens attached to nitrogen so that a trigonal pyramidal shape is achieved. But what forces drive the formation of a tetrahedral shape for methane? Why doesn't it assume a flat plane? Wouldn't that create the greatest distance among the hydrogens?(28 votes)
- If it assumed a flat plane, then the angles between the Hs would be 360°/4 = 90°.
A tetrahedral shape allows the angles between the Hs to be 109.5°, greater than the 90° that a flat plane would allow.(57 votes)
- At7:45, how do you know that the lone pairs on oxygen are adjacent to each other, instead of having one above and one below and the hydrogens forming a 180 degree angle with the oxygen?(28 votes)
- In compounds such as Methane, seen drawn at2:57, all of the 4 bonding electron clouds are a 109.5 degrees from each other. Since water also has 4 electron clouds, it will take the same shape, except with unbonding electron clouds taking the place of two of those bonding electron clouds. So the angle between the hydrogen atoms will always be ~109.5 degrees (104.45, to be presice, due to reasons he stated) and the structure will always be bent/angular, you'll just be looking at if from a different angle depending on where the lone pairs were placed.(10 votes)
- Why do lone pair of electrons occupy more space than bond pairs?(11 votes)
- The lone pair of electrons are situated more closer to the central atom than the bonding electron. Hence, they have more repulsion and thus, occupy more space than bond pairs.(14 votes)
- At9:22and two previous other times, how do you calculate the angles between the hydrogens?(14 votes)
- So if the center has electrons that aren't bond to any other atom is also counted as a electron cloud.(0 votes)
- how is trigonal pyramidal different from tetrahedral? A tetrahedron IS a triangular pyramid.(10 votes)
- A tetrahedron is a triangular pyramid but a triangular pyramid is not necessarily a tetrahedron(the regular polygon), so you're right but by using both terms it gives you just that little bit more precision in describing the shapes.(8 votes)
- How do you predict how much the lone pairs of electrons will push the hydrogens in water? Like, how do they know that the Lone pairs reduce the angle from 107º to 104.5º and not to 100º or 105.5º?(1 vote)
- They can measure the bond angles by techniques like X-ray spectroscopy.(8 votes)
- what is the basic concepT to understand the VSEPR THEORY....??(1 vote)
- The basic concept is that electrons repel each other and will try to get as far away from each other as possible.(8 votes)
- At3:50, how is it that 109.5 separates all the clouds better on a single plane compared to 90?(2 votes)
- But it isn't on a single plane, it's in 3D space. 109.5 degrees just happens to be the maximum angle when you have 4 equal groups.
If you need proof then Jay did do a mathy video that proves this on here, might be easier just accept it for what it is.(5 votes)
- Bond angles are given at 2;40,, at7:00and at9:11but there is no mention of how they are determined/calculated or their importance to a given molecule's structure. In fact, in over an hour of VSEPR there is no mention whatsoever of why a molecule's shape is important.... does anyone know?(2 votes)
- Molecular shape is used in determining the stability of the molecule and how it can bond with other molecules; this becomes more relevant in the section on hybrid orbitals and hybridization.(4 votes)
- Why is 109.5 the appropriate degree for tetrahedral?(2 votes)
- Because in 3D that is the largest angle that can result from 4 atoms bonded to the central atom.
There's a proof video in this playlist but it isn't the most convincing.
Let's figure out the shape of the methane molecule using VSEPR theory. So the first thing that you do is draw a dot structure to show it the valence electrons. So for methane, carbon is in group four. So 4 valence electrons. Hydrogen is in group one, and I have four of them, so 1 times 4 is 4, plus 4 is 8 valence electrons that we need to show in our dot structure. Carbon goes in the center and carbon is bonded to 4 hydrogens, so I can go ahead and put my hydrogens in there like that. And this is a very simple dot structure. We've already shown all 8 of our valence electrons. Let me go ahead and highlight those here. 2, 4, six, and 8. So carbon has an octet and we are done. The next thing we need to do is count the number of electron clouds that surround our central atom. So remember, electron clouds are regions of electron density, all right? So we can think about these bonding electrons here as being electron clouds. So that's one electron cloud. Here's another one down here. And then here's one. And then finally, here's another one. So we have four electron clouds surrounding our central atom. The next step is to predict the geometry of your electron clouds around your central atom. And so VSEPR theory tells us that those valence electrons are going to repel each other since they are negatively charged. And therefore, they're going to try to get as far away from each other as they can in space. And when you have four electron clouds, the electron clouds are farthest away from each other if they point towards the course of a tetrahedron, which is a four sided figure. So let me go ahead and draw the molecule here, draw the methane molecule. I'm going to attempt to show it in a tetrahedral geometry. And then I'll actually show you what a tetrahedron looks like here. So here's a quick sketch of what the molecule sort of looks like. And let me go ahead and draw tetrahedron over here so you can get a little bit better idea of the shape, all right? So four sided figure. And so there you go. Something like that. So you could think about the corners of your tetrahedron as being approximately where your hydrogens are and that just gives you a little bit better visual picture of that tetrahedron, that four sided figure here. And so we've created the geometry of the electron clouds around our central atom. And in step four, we ignore any lone pairs around our central atom, which we have none this time. And so therefore, the geometry the molecule is the same as the geometry of our electron pairs. So we can say that methane is a tetrahedral molecule like that. All right, in terms of bond angles. So our goal now is to figure out what the bond angles are in a tetrahedral molecule. Turns out to be 109.5 degrees in space. So that's having those bonding electrons as far away from each other as they possibly can using VSEPR theory. So 109.5 degrees turns out to be the ideal bond angle for a tetrahedral molecule. Let's go ahead and do another one. Let's look at ammonia. So we have NH3. First thing we need to do is draw the dot structure. So we start by finding our valence electrons, nitrogen in group five. So 5 valence electrons. Hydrogen in group one, and I have three of them. So 1 times 3 plus 5 is 8. So once again, we have 8 valence electrons to worry about. We put nitrogen in the center and we know nitrogen has bonded to 3 hydrogens, so we go ahead and put our 3 hydrogens in there like that. Let's see how many valence electrons we've used up so far. 2, 4, and 6. So 8 minus 6 is 2 valence electrons left. We can't put them on our terminal atoms, because the hydrogens are already surrounded by two electrons. So we go ahead and put those two valence electrons on our central atom, which is our nitrogen like that. And so now we've gone ahead and represented those two valence electrons. So we have all eight valence electrons shown for our dot structure. All right, we go back up here to our steps to remind us what to do after we've drawn our dot structure. And we can see that now we're going to think about the electron clouds that surrounded the central atom. So regions of electron density. And let's go ahead and find those. So I can see that I have these bonding electrons. That's a region of electron density, so that's an electron cloud. Here's another one. So that's two. Here's another one. So that's three. And this lone pair of electrons, this non-bonding pair of electrons is also going to be counted as an electron cloud. It's a region of electron density too. And so once again we have four regions of electron density. When you're thinking about the geometry of those electron clouds, those four electron clouds are going to, once again, try to point towards the corners of a tetrahedron. So we can kind of sketch out the ammonia molecule. And we can draw the base the same way we did before, with our three hydrogens right here. And then we're going to go ahead and put our lone pair of electrons right up here. And so again, it's an attempt to show the electron clouds in a tetrahedral geometry. Let's go back up here and look at our steps again. So in step three, we predicted the geometry electron clouds are going to attempt to be in a tetrahedron shape around our central atom. But when we're actually talking about the geometry or shape of the molecule, we're going to ignore any lone pairs when we predict the geometry of the molecule. So when we look at the ammonia molecule, we're going to ignore that lone pair of electrons on top of the nitrogen and we're just going to focus in on the bottom part for the shape here. And so when we do that, we get something that looks like a little squat pyramid here. So if I'm [? ignoring ?] that lone pair of electrons up there at the top, it's going to look something like that for the shape. And we call this trigonal pyramidal. So this is a trigonal pyramidal shape. So even though the electron clouds are attempting to be in a tetrahedron fashion, the shape is more trigonal pyramidal because we ignore any lone pairs of electrons. In terms of a bond angle, this lone pair of electrons on the nitrogen actually occupies more space. These non-bonding electrons occupy a little more space than bonding electrons. And because of that, those non-bonding electrons are going to repel these bonding electrons. I'm going to go ahead and put them in blue here just as an example. Repel these a little bit more than in the previous example that we saw. And that's actually going to make the bond angle a little bit smaller than the ideal bonding angle we saw before for 109.5 for a tetrahedral arrangement of electron clouds. And so it turns out that this bond angle between the atoms, the hydrogen nitrogen hydrogen bond angle gets a little bit smaller than 109.5. So it actually gets smaller to approximately 107 degrees here for a trigonal pyramidal situation. All right, let's do one more. Let's go ahead and do the water molecule. All right, so we have H2O. And to follow our steps, we know that hydrogen's in group one. I have two of them. And we know that oxygen is in group six. So 6 plus 2 is once again 8 valence electrons to represent for our dot structure. And we put oxygen in the center. Oxygen is bonded to two hydrogens, so we go ahead and draw those in there. And let's see, how many valence electrons have we represented so far? That's 2, that's 4. So 8 minus 4 is 4 valence electrons left. We first think about putting them on our terminal atoms, but those are our hydrogens, so they're already happy with two electrons. So we go ahead and put those four valence electrons on our central atom, which is our oxygen. And four valence electrons means two lone pairs of electrons now. And so now we've represented all eight valence electrons for water. Our next step is, of course, to count how many electron clouds we have around our central atom. So once again, we could think about these bonding electrons as being an electron cloud. These bonding electrons as being electron cloud, these non-bonding electrons as lone pairs in an electron cloud, and same thing for these non-bonding electrons, electron cloud over there as well. And so once again we have four electron clouds. And those four electron clouds are going to attempt to be in a tetrahedral arrangement around that central atom. Using VSEPR theory, they're going to repel each other and get as far away from each other as they possibly can. Let me go and redraw the molecule here. So let's go ahead. And we have our water molecule and we have our lone pairs of electrons like that. And in this case we have two lone pairs of electrons. Remember, lone pairs or non-bonding electrons take up a little bit more space than bonding electrons. And therefore, they're going to repel these electrons right in here a little bit more. And that's going to make our bond angle even smaller than before. So it's going to be even smaller than 107 degrees. And so this bond angle right here, you'll see it listed as approximately 104.5 degrees, or some textbooks will say 105 degrees. So that's approximately what it is for this. In terms of the geometry of the molecule. So the geometry of the electron clouds are attempting to be, once again, a tetrahedral fashion, but the geometry of the molecule is different because you ignore lone pairs of electrons. And so when you look at the shape, if you look at the shape of this-- I'll go ahead and draw the shape over here. If you're ignoring lone pairs of electrons, it looks like that. And we've seen that shape before. That's bent or angular. So we say that the geometry of the water molecule is bent or angular with an approximately 104.5 degree bond angle. So those are a couple examples of four electron clouds and how to figure out the geometry while also thinking about the bond angles.