VSEPR for 5 electron clouds (part 1)
In this video, we apply VSEPR theory to molecules and ions with five groups or “clouds” of electrons around the central atom. To minimize repulsions, five electron clouds will always adopt a trigonal bipyramidal electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be trigonal bipyramidal (no lone pairs), seesaw (one lone pair), T-shaped (two lone pairs), or linear (three lone pairs). This video focuses on the first two molecular geometries, trigonal bipyramidal and seesaw. Created by Jay.
Want to join the conversation?
- Why is the bond angle between the equatorial lone pair and the axial fluorines still 90 degrees? Shouldn't the lone pair repel the axial fluorines so that the angles between axial fluorines and the equatorial lone pair be more than 90 degrees, therefore causing the angles between the equatorial fluorines and the axial fluorines to be less than 90 degrees?(28 votes)
- Sharp thinking! Those are the theoretical bond angles.
The lone pair repels all the bond pairs and does just as you predicted.
The F-S-F bond angle between the equatorial fluorines is reduced from 120 ° to 102 °.
The F-S-F bond angle between the axial fluorines is reduced from 180 ° to 173 °.
The F-S-F bond angle between the axial fluorines and the lone pair is increased from 90 ° to 93.5 °.
The F-S-F bond angle between the equatorial and axial fluorines is reduced from 90 ° to 86.5 °.(51 votes)
- Is the trigonal bipyramidal geometry just observed behaviour or is there like a mathematical proof, that this configuration minimises electron-electron repulsion?(26 votes)
- It's both. In these exercises we are trying to make our best predictions following some basic rules. Scientific experiments will test whether our predictions were correct or not. If they are (in most cases they are) than all right! If they aren't, we want to know why is that. There are many ways to find out spatial distribution of atoms in a molecule.(21 votes)
- why are the lone pairs of electrons around chlorine not counted as electron clouds?(7 votes)
- I think it is because he is talking from the point of view of the CENTRAL atom. It is the electron clouds immediately around phosphorus (the central atom) that gives PCl5's shape as a whole, according to VSEPR model. In other words, it will be useless to analyze PCl5 shape from the point of view of each individual chlorine atoms.(13 votes)
Why is it okay for Phosphorus to have an expanded octet?
What is the relation between period 3 and the expanded octet of Phosphorus?
If possible please give me a link of the Khan Academy video for my question's answer....
Thank you.!(3 votes)
- You will have to refer to a concept called VBT (Valence Bond Theory)(5 votes)
- At12:23Why is it said that the structure on the left is the right one? ALSO why is it that we have only two resonance structures for the molecule at12:15?(3 votes)
- These are isomers, not resonance structures.
The five orbitals have a trigonal bipyramidal geometry.
There are only two places where the lone pair can go — the axial ot the equatorial location.
The see-saw geometry is more stable because the lone pair in the axial location has less total repulsion from the other electrons.(5 votes)
- Why wouldn't the bond angles on SF4 be less than ideal if the lone pair repels the bonding pairs?(3 votes)
- Yes, that's true: the bond angles are "less than ideal" but, to be honest, that is the ideal arrangement for a molecule with a structure like that.
The structure of SF4 is actually a "seesaw" which is a variation of "trigonal bipyramidal" because there is this extra pair of electrons. The bond angles are also smaller(4 votes)
- In the molecular structure, does the Sulfur, the central atom of SF4, has four electrons that is bonded to F?
Is that why we can put two more valence electrons when there are left overs?
So, we can put up to four more valence electrons in a central atom?(4 votes)
- Sulfur is in the 3rd period, meaning it has a d-orbital to store up to 10 extra electrons. In general, the octet rule is less important for the 3rd period down because of these sorts of exceptions(1 vote)
- How do you know when the octet rule applies? is it just when the formal charge isn't 0?(1 vote)
- The octet rule applies for atoms in the second row of the periodic table (most notably: B, C, N, O, F, Ne). The first row (H and He) only has 1s orbitals available, so can only hold a maximum of 2 electrons. The beginning of the second row (Li, Be) often loses electrons to look more like the first row (i.e. 2 electrons). From the third row onwards, atoms can sometimes have more than an octet ("expanded octet"), since they now have d orbitals available, too.
Formal charges do not impact whether an atom follows the 'octet rule' or not.(6 votes)
- At9:30, Why do non bonding electrons take up more space than bonding electrons?(1 vote)
- Bonding electrons have to spend most of their tine between two nuclei.
Nonbonding electrons are attracted to a nucleus only from one side, so they are free to wander further away.(5 votes)
- Isn't the NH3 molecule also a trigonal bipyramidal?(1 vote)
- No, NH3 is going to have a tetrahedral electron domain geometry because it has four areas of electrons: a single bond to each hydrogen (three in total) and a lone pair of electrons. This means its molecular geometry will be trigonal pyramidal.(4 votes)
Let's use VSEPR theory to predict the structure of this molecule-- so phosphorus pentachloride. So the first thing we need to do is draw a dot structure to show our valence electrons. We find phosphorus in Group 5. So 5 valence electrons. Chlorine in Group 7. So 7 valence electrons, and I have 5 of them. So 7 times 5 is 35. Plus 5 gives us a total of 40 valence electrons that we need to show in our dot structure. So phosphorus goes in the center because it is not as electronegative as chlorine. And we have five chlorines. So we go ahead and put our five chlorines around our central phosphorus atoms like that. If we see how many valence electrons we've drawn so far, this would to be 2, 4, 6, 8, and 10. So 40 minus 10 gives us 30 valence electrons left over. And remember, you start putting those leftover electrons on your terminal atom. So we're going to put those on the chlorines. Each chlorine's going to follow the octet rule. So that means each chlorine needs 6 more electrons. Now, each chlorine is surrounded by 8 valence electrons like that. So if I'm adding 6 more electrons to 5 atoms, 6 times 5 is 30. So I have now represented all of my valence electrons on my dot structure. Notice that phosphorus is exceeding the octet rule here. There are 10 valence electrons around phosphorus. And it's OK for phosphorus to do that because it's in Period 3 on the periodic table. I like to think about formal charge. And so if you assign a formal charge to phosphorus, you'll see it has a formal charge of 0. And that helps to explain-- for me, anyway-- the resulting dot structure. Now, step two. We're going to count the number of electron clouds that surround our central atom. Remember, an electron cloud is just a region of electron density. So I could think about these bonding electrons in here as a region of electron density around my central atom. I could think about these bonding electrons, too. So here's another electron cloud. And you can see we have a total of five electron clouds around our central atom. The next step is to predict the geometry of the electron clouds. Those valence shell electrons are going to repel each other. All right. So that's a VSEPR theory-- Valence Shell Electron Pair Repulsion. Since they're all negatively charged, they're going to repel and try to get as far away from each other as they possibly can in space. When you have five electron pairs, it turns out the furthest they can get away from each other in space is a shape called a trigonal bipyramidal shape. So let me see if I can draw our molecule in that shape. We're going to have our phosphorus in the center, and we're going to have three chlorines on the same plane. So let me attempt to show three chlorines on the same plane here. These are called the equatorial positions because they're kind of along the equator, if you will. So three chlorines in the same plane, one chlorine above the plane, and one chlorine below the plane. Those are called axial positions. All right. So there's a quick sketch. Let me see if I can draw a slightly better shape of a trigonal bipyramidal shape here. So let me see if I can draw one over here so you can see what it looks like a little bit better. So we could have one pyramid looking something like that. And then, down here, let's see if we can draw another pyramid in here like that. So that's a rough drawing, but we're trying to go for a trigonal bipyramidal shape here. So let's focus in on those chlorines that are on the same plane first. If I'm looking at these three chlorines and I go over here to my trigonal bipyramidal shape, you could think about those three chlorines as being at these corners here. So it's a little bit easier to see. They're in the same plane. So those are the equatorial chlorines. When I think about the bond angle for those-- so those chlorines being in the same plane, you have these three bond angles here. And so when we did trigonal planar, we talked about 360 degrees divided by 3-- giving us a bond angle of 120 degrees. So you could think about that as being a bond angle of 120. All right. So same idea. Those bonding electrons are going to repel each other. When we focus in on our axial chlorines-- so this one up here and this one down here. You could think about those as being here and here on your trigonal bipyramidal shape like that. And if you draw the axis, if you draw a line down this way connecting those, it's easy to see those are 180 degrees from each other. So you could think about a bond angle of 180 degrees between your chlorines like that. And then, finally, if we think about the bond angle between, let's say, this axial chlorine up here at the top and then one of these green chlorines right here, I think it's a little bit easier to see that's 90 degrees here. So this bond angle right here would be 90 degrees. And so those are your three ideal bond angles for a trigonal bipyramidal situation here. It's important to understand this trigonal bipyramidal shape because all of the five electron cloud drawings that we're going to do are going to have the electron clouds want to take this shape. So it's important to understand those positions. For step four, ignore any lone pairs and predict the geometry of the molecule. Well, there are no lone pairs on our central phosphorus. So the electron clouds take a trigonal bipyramidal shape and so does the molecule. Let's go ahead and do another example. Sulfur tetrafluoride here. So we're going to start by drawing the dot structure, and we need to count our valence electrons, of course. So sulfur's in Group 6, so 6 valence electrons. Fluorine is in Group 7. So 7 valence electrons. I have 4 of them. 7 times 4 is 28. 28 plus 6 is 34 valence electrons. We know sulfur is going to go in the center because fluorine is much more electronegative. We put sulfur in the center here. We know sulfur is bonded to 4 fluorines. So we put our fluorines around like that. And let's see how many valence electrons we've shown so far-- 2, 4, 6, and 8. So 34 minus 8 gives us 26 valence electrons we still need to account for on our dot structure. We're going to start by putting those leftover electrons on our terminal atoms, which are our fluorines. Fluorine's going to have an octet of electrons around it. Therefore, each fluorine needs six, since each fluorine already has two around it. So we go ahead and put 6 valence electrons around each one of our fluorine atoms. All right. So we are showing 6 more valence electrons on 4 atoms. 6 times 4 is 24. So 26 minus 24 gives us 2 leftover valence electrons. And remember your rules for drawing dot structures. When you get some leftover electrons, you're going to go ahead and put them on your central atom now. So we have a lone pair of electrons on our sulfur. And by adding that lone pair of electrons to our sulfur, the sulfur now exceeds the octet rule. But once again, it's OK for sulfur to have an expanded valence shell. It's in Period 3 on our periodic table. And once again, I like to think about formal charge. And if you assign a formal charge to that sulfur, it has a formal charge of 0. So that just helps me understand these dot structures a little bit better. So we've drawn our dot structure. Let's go back up and remind ourselves of the next step here. So once you complete step one, next is the electron cloud step. So how many electron clouds do you have surrounding your central atom? So we go back down, and we look at our electron clouds that surround our central atom. Here's the regions of electron density. So we know that these bombing electrons here, that would be one electron cloud. Same with these bonding electrons. Same with these bonding electrons. And same with these bonding electrons. And then we have a lone pair of electrons on our sulfur. Well, that's also a region of electron density surrounding our central atom. So that lone pair you could think of as being an electron cloud as well. And so we have five electron clouds. So just like in the previous example. And when you have five electron clouds, those electron clouds are going to try to adopt a trigonal bipyramidal shape-- just like we saw, again, in the previous example. So let's go ahead and draw two possible versions of the dot structure for this molecule. All right. So I'm going to draw one right here. And for this first version, I'm going to show the lone pair of electrons on the sulfur in the equatorial position here. So I'm going to put the lone pair of electrons right here, equatorial here. And so that means there are two fluorines also equatorial. And then that means that there's one fluorine here, axial. Another fluorine here, axial. So that is one possible dot structure. The other possibility would be, of course, to put to the lone pair of electrons in the axial position. So if we do that, we would have a sulfur bonded to three fluorines. Those would be the equatorial fluorines. And then, we would have a lone pair of electrons. Let's just put it right here in the axial position. And then, another fluorine in the axial position. So here are our two possibilities. So let's see if we can analyze this structure. Now, when you have lone pairs of electrons in your dot structure, lone pairs take up more space. Or non-bonding electrons-- I should say-- take up more space than bonding electrons. And so since they take up more space, they're going to repel a little bit more. And so that means that when you're trying to figure out valence shell electron pair repulsion, the lone pairs of electrons are more important to focus on in terms of where you're going to put them. Let's focus in on those lone pairs of electrons, and let's think about how they're going to repel the other electrons in these two dot structures. Let's look at the left here where we had the lone pair of electrons in the equatorial position here. And if you're thinking about how they're interacting with, let's say, these bonding electrons in the same plane here, this is about 120 degrees between the bonding electrons and the non-bonding electrons. And it turns out that 120 degrees is not as important in terms of repelling as, say, something like 9 degrees. You tend to ignore the 120 degree interactions when you're analyzing these structures. However, a 90-degree angle between a bonding pair and a non-bonding pair-- and we had that example for-- let me go ahead and show you right here. So let's think about this lone pair of electrons repelling these bonding electrons. So in the axial position. Well, these two are only 90 degrees away. So remember, 9 degrees, of course, being closer, you're going to get more repulsion from this interaction than in the previous interactions. So we're going to focus in on the 90-degree interactions here. Those bonding electrons and non-bonding electrons repel each other. And you have one possibility with the axial fluorine. You also have another possibility with this axial fluorine. Essentially, you have a lone pair of electrons, 90 degrees, from two pairs of bonding electrons from the example on the left. And of course, that's going to destabilize it somewhat. But let's compare this dot structure with the one on the right now. So we have our lone pair of electrons in the axial position this time. And you can see that we have three fluorines in the equatorial positions. So you have these bonding electrons in the equatorial position, which means that that lone pair of electrons is 90 degrees to all three of those. And so that, of course, is going to cause some serious repulsion, so 90 degrees to 3. In the example on the right, you have these three interactions-- 90 degrees. An example on the left, you have only two of these. The goal, of course, is to minimize electron pair repulsion. So VSEPR theory actually predicts that this dot structure on the left is the correct one. You're going to see-- in the next video-- that non-bonding electrons are placed in equatorial positions in trigonal bipyramids to minimize electron pair repulsion. So just think about putting your lone pairs of electrons in the equatorial position. So the structure on the left wins. Let's go ahead and redraw that so we can analyze it a little bit better. All right. So I have my sulfur in the center here. Let's go ahead and change colors. So I'm going to put the sulfur in the center. I have my fluorine in a plane. Another fluorine in the plane. My lone pair of electrons in a plane. All right. Those are my equatorial ones. I have a fluorine this way. And I have a fluorine that way. So when you're looking at bond angles, of course, between this sulfur fluorine bond angle-- the ideal bond angle anyway-- would be 120 degrees. So we can say and we would expect it to be 120 degrees. If you're talking about this axial fluorine and this equatorial one, we would expect that to be 90 degrees. And then, finally, between the two axial fluorines-- so this bond angle back here would, of course, be 180 degrees. OK. So we've done a lot of talking, and we still haven't even talked about the final name for the shape of this molecule. So let's go back up here and look at our rules really fast. So we've done a lot of work to predict the geometry of the electron clouds around the central atom and draw it. And finally, we get to predict the shape of the molecule. And we do that by ignoring any lone pairs. So let's go ahead and do that. So we're going to ignore the lone pair of electrons on the sulfur when we're talking about the shape. So if we ignore the lone pair and we actually turn this molecule on its side-- so let's go ahead and do that. We're going to put our sulfur here. If we turn it on its side, the axial fluorines would now be horizontal. Now, it's horizontal like that. So I'll go ahead and put in my-- so these are the two fluorines that used to be axial there. And my two fluorines that were equatorial, they would look something like this. And it helps if you actually build this molecule with a MolyMod set. So that would be what the molecule kind of looks like here, and we call this a seesaw shape. So this is a seeshaw shape or geometry. And let's think about why. So if you've ever been on a playground and used a seesaw-- I'm going to draw a little kid here on one side of our seesaw like that. And so if the little kid puts his weight on this side, of course, this side of the seesaw salt would go down. And then, this side of the seesaw would go up. So just a little bit of intuition as to why you would call this a seesaw shape. All right. So I think we'll have to stop there. In the next video, we'll do two more examples of molecules and ions that have five electron clouds.