In this video, we apply VSEPR theory to molecules and ions with six groups or “clouds” of electrons around the central atom. To minimize repulsions, six electron clouds will always adopt a octahedral electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be octahedral (no lone pairs), square pyramidal (one lone pair), or square planer (two lone pairs). Created by Jay.
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- At8:30, why doesn't the lone pair repel the fluorine atoms to create a bond angle less than 90 degrees?(13 votes)
- It does, the prediction made in the video was wrong. In BrF₅, the four F atoms that are predicted to be in the same plane with each other actually have bond angles of 89.5° with each other and have a bond angle of 84.8° with the remaining atom of F (the F that is linear with the lone pair and the Br).
So, you are correct. Unfortunately, some resources on VESPR don't take into account the fact that lone pairs take up more space than bond and "push" the bonds out of ideal geometries.(19 votes)
- Can I place one lone pair in the axial side and one at an equitorial side for XeF4 ?(6 votes)
- No. The 3D structure is unstable in that conformation. The lone pairs are more negatively charged than the bonds, so the lone pairs will try to be as far apart as possible. This would place them either both axial, or opposite equatorial. They cannot be both axial and equatorial.(12 votes)
- the lone pair must be on equitorial . but in BrF5 molecule it is on axial line?i(5 votes)
- Octahedral or square bipyramidal has a 90* bond angle between the axial and equitorial atoms both. Regarless of how you spin the molecule it looks the same from all sides. Just pick off an atom to get Square Pyramidal. (Axial and Equatorial don't really axis for Octahedral - play with a model kit)
Trigonal bipyramidal DOES have axial (90*) and equatorial (120*) bond angles. If you remove an equatorial, you still have 2 atoms forming tight 90* bonds to both the axials (Electron repulsion = yikes). If you remove an axial, you have the three equatorials with no 90* bonds formed to the axial thats removed (less electron repulsion). Try it on a model kit(6 votes)
- How come the lone pairs aren't assigned an equatorial position in this example? In the previous videos it was mentioned that lone pairs should be assigned an equatorial position? :/(5 votes)
- It may be because all the Fluorines are the same distance from each other here so it doesn't matter where the lone pairs are placed in relation to the Fluorines. When they have their free choice of where to go, they choose positions that are furtherest away from each other as they repel each other.(3 votes)
- With 6 electron clouds, can't there be cases with 3 lone pairs and 4 lone pairs as well? If so, what are the shapes of those?(3 votes)
- Can't there be 3 THREE lone pairs and 3 bond pairs ??
What will be the structure then??
Example -> XeF3(-1)(2 votes)
- A molecule can be formed using 3 lone pairs and 3 bonds but the central atom cannot have 3 lone pairs and 3 bonds.
The reason being is because then the electron the central atom would have would be ((1/2) * bonds = 3) + (1 for each lone pair electron (2 * 3 = 6)) = 9.
So in short having 3 pairs of shared electrons and 3 pairs of lone electrons would mean it would have 9 valence electrons which is invalid.
Hope this helps,
- Convenient Colleague(2 votes)
- At2:00how do we figure out that the electron cloud geometry was a regular octahedron? Is it worked out experimentally or is it something that we could deduce just by the number of electron density regions?(2 votes)
- It was determined experimentally. This was made possible due to development of advanced technology such as spectroscopy.(2 votes)
- In drawing the structures, it is always mentioned "[...] is following the octet rule". How do I know which atom follows the octet rule? Don't they all?(1 vote)
- Only the atoms from C to Si must follow the octet rule.
Atoms after Si often follow the octet rule, but they may "expand their octets" and disobey the octet rule.(4 votes)
- At6:52Mr Jay says it doesn't really matter where we put the lone pair of electrons.... But surely the axial position is preferable because, although it is 90° from the equatorial regions, it gives 180° from the other axial electron density region - whereas the equatorial position is 90° from everything...?(2 votes)
- See, just look at the figure of the octahedron very carefully. If you rotate the octahedron(or just tilt your head and see), the positions which seemed to be equatorial will become axial and the seemingly axial positions will become equatorial. The point Jay is making is that every position is exactly same-90 degree angle with 4 neighboring positions and 180 degrees angle with 1 position. It might feel a bit counter-intuitive but try drawing the octahedral structure of the first compound, choose any one equatorial position and look carefully whether it can also act as an axial position. The equatorial position is in the same plane as three other positions;i.e.-forms a rectangle;and two positions project upward and downward. Hence, every position is same, unlike the situation with 5 electron cloud species. That's why the position of the lone pair does not matter in the second example.(1 vote)
- what is equatorial and axial?(1 vote)
- Think of the atom like the earth. Equatorial refers to the electrons that are on the earth's equator. Axial refers to the electrons that are at the north and south poles - on the earth's axis.(3 votes)
In this video, we're going to apply VSEPR theory to 6 electron clouds. So if our goal is to find the shape of the sulfur hexafluoride molecule, once again we start with our dot structure. So sulfur is in group 6 on the periodic table, so 6 valence electrons. Fluorine is in group 7, so 7 valence electrons, but I have 6 of them. So 7 times 6 gives me 42, and 42 plus 6 gives me 48 valence electrons that we need to show in our dot structure. Sulfur goes in the center, so we go ahead and put sulfur there. And we surround sulfur with 6 fluorines. So let me go ahead and put in those 6 fluorines surrounding our sulfur. Our next step is to see how many valence electrons that we've shown so far. So I go and highlight those-- 2, 4, 6, 8, 10, and 12. So 48 minus 12 gives me 36 valence electrons left over, which we put on our terminal atoms, which are our fluorines. So fluorine is going to follow the octet rule. And since each fluorine is already surrounded by 2 electrons, we're going to give each fluorine 6 more. So by giving each fluorine 6 more, now each fluorine has an octet of electrons around it. So if I'm adding 6 electrons to 6 atoms, 6 times 6 is 36. And so therefore, I have now represented of all my valence electrons. And we're done with our dot structure. We can move on to step two and count the number of electron clouds surrounding our central atom, so regions of electron density. So these bonding electrons here, that's a region of electron density. And I can just keep on going all the way around. So all of these bonding electrons surrounding our sulfur are regions of electron density. Therefore, we consider them to be electron clouds. VSEPR theory says that these valence electrons are all negatively charged, and therefore, they're all going to repel each other and try to get as far away from each other in space as they possibly can. And so when you have 6 electron clouds, they're going to point towards the corners of a regular octahedron to try to get as far away from each other as they can. So an octahedron with 8 faces on it. So let me see if I can sketch in an octahedron here. So let's see if we can do it. It's a little bit tricky to draw. So if we consider our sulfur to be at the center right here-- let's go ahead and put a point up here and then start connecting some lines. So this is sort of what it looks like. So let's do that and then a point down here as well. And so we connect those lines. And once again, this is just a rough sketch of an octahedron. Something like that. So if you think about where your fluorines are-- right there. Here's a fluorine right here. There's a fluorine right here. So at these corners, you could think about a fluorine being there like that. So that's our octahedron. So that's step three. The geometry of the electron clouds around the central atom, they occupy an octahedral geometry. Step four, ignore any lone pairs in your central atom and predict the geometry of the molecule. Well, since we have no lone pairs on our central sulfur, the geometry of the molecule is the same as the geometry of the electron clouds. And so therefore, we can say that sulfur hexafluoride is an octahedral molecule. So let's go ahead and write octahedral here. In terms of bond angles, let's analyze our drawing a little bit more here. So if I look at this top fluorine and I go straight down like an axis to that other fluorine, we would expect one of the ideal bond angles to be 180 degrees for this octahedron here. And the other ideal bond angles would be 90 degrees. So if I think about the angle that the axis I just drew makes with this one right here, so that's 90 degrees as well. And again, anywhere you look, you're also going to get 90 degrees. Let me go ahead and change colors here, and we can look at another bond angle in here. So this bond angle, that would also be 90 degrees. So for an octahedral, all 6 positions-- we have 6 fluorines occupying the 6 positions-- are equivalents. They are identical, which means no axial or equatorial groups in an octahedral arrangement. And that makes our life much easier, because in the videos on 5 electron clouds, we had to think about the axial and equatorial groups. Let's do one for bromine pentafluoride here, so BrF5. So valence electrons, bromine has 7. It's in group 7. Fluorine is also in group 7, and I have 5 fluorine. So 7 times 5 gives me 35. 35 plus 7 gives me 42 valence electrons. Bromine goes in the center, and bromine is bonded to 5 fluorines. So I can go ahead and put those 5 fluorines around our central atom. We have represented, let's see, 2, 4, 6, 8, and 10 valence electrons so far. 42 minus 10 is, of course, 32 valence electrons. And we're going to start putting those leftover electrons on our terminal atoms, which are our fluorines. So once again, we're going to give each fluorine an octet. So we're going to put 6 more valence electrons around each of our fluorine atoms. And so we're putting 6 more electrons around 5 atoms. So 6 times 5 is 30. So 32 minus 30 gives me 2 valence electrons left over. And whenever you have valence electrons left over after assigning them to your terminal atoms, you put them on your central atom. And so there's going to be a lone pair of electrons on our central bromine like that. So we've drawn our dot structure. Let's go back up here and look at our steps again. So after drawing our dot structure, we next count the number of electron clouds that surround our central atom and then predict the geometry of those electron clouds. And so if we look at our central bromine here, let's see how many electron clouds. Well, we would have these bonding electrons, a region of electron density, these bonding electrons, these bonding electrons. And we keep on going around here. So those are all electron clouds. So that's 5. And then remember, these non-bonding electrons, this lone pair of electrons, is also a region of electron density. And so we have 6 electron clouds. And so we just saw in the previous example, when you have 6 electron clouds, the electron clouds are going to want to point towards the corners of a regular octahedron. So you're going to get an octahedral geometry for your electron clouds. Let's think about this one, though. Where will we put those that lone pair of electrons in an octahedron. Well, since all 6 positions are identical, it doesn't really matter which one you put that lone pair of electrons in. And so let me see if I can just go ahead and sketch out the shape really fast. So if I were to draw a bromine right here, I'm going to put a fluorine going in this direction, another fluorine going back, this one coming out a little bit, and this one going away. And then I'm going to put a fluorine going this way, and then I'm going to put the lone pair of electrons right down here. Again, it didn't really matter which one I chose since they're all identical. I just chose it this way because it's a little bit easier to see the geometry. Because when you're looking at the geometry of the molecule, you ignore any lone pairs of electrons on your central atom. And so if we ignore that lone pair of electrons now, and we look at the shape-- so let's see if we can connect these dots here. So we're just going to connect this to look at a shape. So we have a square base here. And if we connect up here to this top fluorine, well, that's kind of a pyramid. So we have a pyramid with a square base. And so we call this square pyramidal. So let's go ahead and write that. This shape is referred to as a square pyramidal shape. And in terms of bond angles, we know our ideal bond angles are going to be 90 degrees. So if we look at that, let's use this green here. So it's just like we talked about before. So that bond angle is 90 degrees. This bond angle in here is 90 degrees. So our ideal bond angles are all 90 degrees for our square pyramidal geometry. Let's do one more example of 6 electron clouds. And this is xenon tetrafluoride. So we need to find our valence electrons. So xenon is in group 8, 8 valence electrons. Fluorine is in group 7. So 7 valence electrons times 4 gives me 28. 28 plus 8 gives me 36 valence electrons. Xenon goes in the center. So we go ahead and put xenon in the center here. And xenon is bonded to 4 fluorines. So we go ahead and put in our 4 fluorines surrounding our xenon. And let's see, we have represented 2, 4, 6, and 8 valence electrons. So 36 minus 8, that would give me 28 valence electrons left over, which we will put on our terminal fluorines here. So each fluorine is going to get an octet. And so we need to put 6 valence electrons on each one of our fluorine atoms. So we are representing 6 more electrons on 4 atoms. So 6 times 4 is 24. So 28 minus 24 gives us 4 valence electrons left over. And we're going to put those on our central atom here, so we're going to put those on the xenon. So we go ahead and add in those 4 electrons in the form of two lone pairs to our central atom. All right. Let's go back up and refresh our memory about what we do after we draw our dot structure. So after you draw your dot structure, you count the number of electron clouds. And then you predict the geometry of those electron clouds, and so let's count our electron clouds for this one. So our regions of electron density, so we can see that these bonding electrons are an electron cloud, same with these bonding electrons, and these over here as well. And in this example, we have 2 lone pairs of electrons, and each one of those is a region of electron density. And so we have a total of 6 electron clouds for this example. So once again, 6 electron clouds, they are going to want to get as far away from each other as they can. So they are going to be in an octahedral arrangement, and so let's see if we can sketch out this molecule again. So if the lone pairs of electrons want to get as far away from each other as they possibly can, we're going to put those lone pairs 180 degrees from each other. So here's one lone pair, and then here's our other lone pair. That's as far away from each other as they can get. And then we're going to put our fluorines in here. So here's one fluorine. Here would be another fluorine. And then we would have two more back here. So when we look at the shape of xenon tetrafluoride, let's see if we can sketch in what the shape would look like here. So remember, when you're predicting the geometry of the molecule, you ignore the lone pairs of electrons. And so that makes it much easier to see that we have a square that is planar. So we call this square planar. So the geometry is square planar. And in terms of ideal bond angles, that would be 90 degrees. So let me go ahead and show that real fast. So in terms of bond angles, everything here would be 90 degrees for our square planar. And so that's how to approach 6 electron clouds. And our first example had 0 lone pairs of electrons around the central atom. Our second example had 1 lone pair. And our third example had 2 lone pairs. And so even though the electron clouds have the same geometry, the actual molecule is said to have a different shape, because you ignore the lone pairs of electrons on your central atom.