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Lesson 7: VSEPR

# VSEPR for 5 electron clouds (part 2)

In this video, we continue exploring VSEPR theory for molecules and ions with five regions or “clouds” of electrons around the central atom, focusing on examples where the molecular geometry is T-shaped (two of the clouds are lone pairs) or linear (three of the clouds are lone pairs). Created by Jay.

## Want to join the conversation?

• For CIF3 my intuition tells me that putting the 2 lone pairs on the north and the south balances the bond more. The 3 Fluorine will be on the equatorial positions with 120 degree angle from each other. I do understand that by putting the lone pairs on the north and south will have a 90 degree angle, which will be closer to affect the Fluorine atoms, which are located on the equatorial position. However, because we have two lone pairs on the opposite sides of each other, it will have the same amount of push to all the Fluorine atoms. Thus will create an equilibrium to the atom by making it more balance. Is the way I am thinking correct? or am I very wrong. Thank you for your help in advance.
(42 votes)
• I totally get what you are saying, and it totally makes sense in terms of balancing each other out. However, one of the tenets of VSEPR theory is that lone pairs oriented themselves as far apart as possible from other lone pairs as well as bonding pairs to minimize their interactions since they are all negatively charged.

When the electron pairs are in the axial positions (north and south), they are always only 90 degrees from every bond. However, when the electron pairs are in the trigonal positions, then they are 120 degrees from each other and 90 degrees from the axial bonds. That means in the latter they are oriented as far apart as possible.
(42 votes)
• Why cant we put double bonds on each side of the centeral atom and leave on lone pair of electron the formal charge will be +1 ?
(12 votes)
• If you had two double bonds the the central Iodine, the Formal Charge on the outside Iodines would each be +1. Since there are two that would induce a net force of +2 on the molecule. The way Sal draws it leaves a Formal Charge of -1 on the Internal Iodine with a 0 on the outside Iodines. Net force on this molecule then is -1. This is the preferred resonance structure
(10 votes)
• In these VSEPR videos, Jay keeps saying "its ok for X element to break the octet rule because its on the 3rd period". I understand why this is ok when the formal charge equals out to be 0. I don't understand why the 3rd period allows this.
(9 votes)
• The general explanation is that elements in the third period and on have empty d orbitals they can use to bond.
(16 votes)
• I tend to get confused with the term equatorial. Is their an easy way you can explain to me what that means? So I can grasp a better picture. Thanks!
(4 votes)
• equatorial is like the equator - a horizontal line either pointing slightly upwards or downwards. often this is confused with axial which refers to straight up and down.
(9 votes)
• And why is it I3-? not I3?
(5 votes)
• The structure will look like: I-I-I
So, each the two iodines on the side have a single bond and the one in the center has two single bonds. The two iodines on the sides have neutral charge because they have 8 electrons but the one in the center now has more electrons than needed (9) which is why it is negatively charged. It is made from I2 + I(-) --> I3(-)
(4 votes)
• why dont we take lone pair above and below the Cl atom
(3 votes)
• The lone pairs are nonbonding, so they occupy more space than the bonding pairs.
If they occupy the axial positions, the three bonding pairs are at angles of 90° to the lone pairs, and there is a certain LP-BP repulsion,
In the equatorial positions, the angles are 120°. Since the repulsions decrease with the square of the distance, the equatorial positions are preferred.
lone pairs will always occupy the equatorial locations if possible.
(7 votes)
• why do we ignore lone pair electrons for final structure in vsepr theory?
(4 votes)
• Because that's how molecular geometry is defined, it's only based on the shape of the bonded atoms.
(3 votes)
• At why can't you put the lone electron clouds on the axial positions? Wouldn't that minimize electron repulsion?
(3 votes)
• If you put the lone pairs of electrons in the axial positions, you will actually have more electron repulsion. The axial positions form a 90 degree angle with the equatorial positions. If the lone pairs were in the axial position, each lone pair will have repulsion with each of the three F's in the equatorial positions (ie total 6). On the other hand, having the lone pairs in the equatorial position means that each of the lone pairs will only experience repulsion twice, once with each F in the axial position (ie 4). Since 6 > 4, there is less repulsion when the lone pairs are the equatorial position. Hope this makes sense!
(3 votes)
• Why don't we place the lone pairs in the axial direction in the structure of ClF3, as the lone pairs will be farthest away from each and will equally exert forces on the bonding electrons thus form a stabilised structure.... I am not able understand it although I understand the case of SF4...Please help.
(4 votes)
• in i3- ion........shouldnt it have 11 electron clouds as it has 9 lone pairs
(2 votes)
• You should only consider the central atom.
(4 votes)

## Video transcript

In the previous video, we used VSEPR theory for 5 electron clouds. In this video, we're just going to do a few more examples of 5 electron clouds. So let's say we want to find the shape of chlorine trifluoride. We start by drawing a dot structure, which means we have to figure out the valence electrons. So for chlorine in group 7, therefore, 7 valence electrons. Fluorine is in group 7, and we have 3 fluorines. So 7 times 3 is 21, plus 7 means we have a total of 28 valence electrons that we need to show in our dot structure. Chlorine's going to go in the center since it is not as electronegative as fluorine. And we know that the chlorine is bonded to 3 fluorines, so we can go ahead and put it in our 3 fluorines in here like that. We have represented 2, 4, and 6 valence electrons so far. So 28 minus 6 means we have 22 valence electrons left. And we know we start by putting those leftover valence electrons on our terminal atoms, which are our fluorine. Fluorine is going to follow the octet rule, and each fluorine already has 2 electrons around it, and so therefore, each fluorine needs 6 more. So when we put 6 more valence electrons around each fluorine, now each fluorine has an octet. We just represented 6 times 3 more valence electrons, so 6 times 3 is 18, and 22 minus 18 is 4. So we're left with 4 valence electrons. And when you have leftover valence electrons after assigning them to your terminal atoms, you're going to give them to your central atoms. So we're going to assign 4 valence electrons to our central atom, and it makes sense to put those in lone pairs. So there's one lone pair, and there's another lone pair on the chlorine. And so now that takes care of all of our valence electrons. In terms of looking at the dot structure and thinking about chlorine there, it actually is exceeding the octet rule. And that's OK for chlorine to exceed the octet rule, because of its position in the third period on the periodic table. I like to think about formal charge. So if you actually assign a formal charge to that chlorine, you'll see it has a formal charge of 0, which to me just helps me understand these dot structures a little bit more. And so we've drawn a dot structure that makes sense, and so we can move on now to the second step, which is where we count the number of electron clouds that surround our central atom. So remember, electron clouds are regions of electron density, which means that bonding and non-bonding electrons would fit into that category. So here we have a bonding pair of electrons occupying an electron cloud. Here's another bonding pair of electrons in an electron cloud. Here's another one. And then we have our lone pairs of electrons. Our non-bonding electrons are still regions of electron density, and so there is an electron cloud, and then our last lone pair there. So a total of 5 electron clouds surrounding our central atom. In step 3, you predict the geometry of those electron clouds around your central atom. And in the previous video, we used VSEPR theory to talk about why 5 electron clouds are going to form a trigonal bipyramidal geometry. So they're going to repel each other as much as they possibly can. It turns out that's a trigonal bipyramidal geometry here. So our electron clouds are going to be in the same geometry here. So the only tricky part for this dot structure is where do you put your lone pairs of electrons? Lone pairs of electrons take up more space. These non-bonding lone pairs of electrons take up more space than bonding electrons, and therefore, they repel more, so where you put them is extremely important for the overall structure of the molecule. In the last video, we talked in a lot of detail about the fact that you're going to put non-bonding electrons, or lone pairs of electrons, in the equatorial position to minimize electron pair repulsion. And so we're going to do the same thing here. We're going to put our lone pairs of electrons equatorial. So let me go ahead and put in my central chlorine atom. I'm going to put a lone pair of electrons in the equatorial here, and then the other lone pair of electrons also equatorial right here. And that means that I have one more spot. I'm going to put one of the fluorines in the last equatorial position. And so that leaves 2 more fluorines, which we're going to put axial, so one fluorine axial here and one fluorine axial here. Again, putting your lone pairs equatorial minimizes electron pair repulsion, and watch the previous video for more details about that. This is actually a more complicated example than what I talk about in the previous video, but the same ideas apply here as well. I could probably spend a whole video talking about just this one molecule, but we don't really have time for that. So when you're talking about 5 electron clouds, just think about putting your lone pairs in the equatorial position. And so now we have the general structure here. Let's go ahead and talk about the final shape. So when you're talking about the final shape, you ignore any lone pairs, and you predict the geometry of the molecule here. So if we ignore lone pairs, let me go ahead and redraw that. So we're going to go ahead and put in our chlorine here. So there's our chlorine. We ignore lone pairs. So we have our 2 axial fluorines, and then we have our equatorial fluorine 90 degrees to our axial fluorines. And so-- well, that's of course, an ideal bond angle. So let's talk about the bond angles here. So if you look at the shape, so it looks kind of like these are linear here, and then this is 90 degrees to that, so you can see kind of a T-shape here. And so we actually call this molecule, the shape, T-shaped, right? Because we're ignoring the lone pair of electrons, and so we see a T-shape right here, so this is a T-shaped geometry. In terms of the ideal bond angles, since it is T-shaped, it makes it simple. So you could think about for your bond angles, this one you would expect to find a 90-degree bond angle for the fluorine-chlorine-fluorine bond angle here, and then 180 degrees on this side. But once again, those are just ideal bond angles. That's what you would predict thinking about a simple T-shape. The actual bond angles you would have to get to experimentally, so T-shaped geometry here. All right. Let's do one more example of 5 electron clouds. So this is the triiodide ion, so I3 with a negative charge here. So we need to draw the dot structures, and we need the valence electrons. Iodine's in group 7, so 7 times 3 gives us 21 valence electrons. But this is an ion, so it has a negative charge. It has an extra electron, so we have to add 1 to that. So 21 plus 1 gives us a total of 22 valence electrons to show in our dot structure. So we have 3 iodines, so we can go ahead and show our 3 iodines bonded together like that. And we've already represented 4 valence electrons, so 2 here and 2 here. So 22 minus 4 means we have 18 valence electrons left. And we start by putting those leftover valence electrons on our terminal atoms, which are our iodines. So we're going to think about our terminal atoms as following the octet rule. Here, and so we're going to put 6 more electrons on each iodine to give each iodine an octet. And so I just represented 12 more valence electrons, so 18 minus 12 gives us 6 valence electrons left over. And those leftover electrons are, of course, going to be assigned to our central atom. And so with 6 valence electrons, you would expect 3 lone pairs of electrons. So let's go ahead and put those last 6 valence electrons around our central iodine in the form of 3 lone pairs. And since this is an ion, we should put this in brackets and put a negative 1 charge outside like that. So this is our dot structure. And if we look at our central iodine, it's once again exceeding our octet. So it's once again OK for iodine to expand its valence shell, because of its position on the periodic table. And if you assign a formal charge to that iodine, you'll see it's a negative 1 formal charge, which, once again, I always like to do just to help me understand what's going on a little bit better here. So this is a dot structure that makes sense, that follows all of our rules. And so let's go back up here to our next step. So number 1, we've already drawn a dot structure to show our valence electrons. In step 2, we count the number of electron clouds that surround our central atom. So let's go ahead and look at our central atom and figure out how many electron clouds surround it. So these bonding electrons are an electron cloud. These bonding electrons are an electron cloud. And our non-bonding electrons, our lone pairs of electrons, are still regions of electron density. And so you can see we have a total of 5 electron clouds for this structure, too. So for 5 electron clouds, that's going to be a trigonal bipyramidal arrangement of our electron clouds around our central atom. So go ahead and put in our central atom here. We think about our electron clouds being in a trigonal bipyramidal arrangement. And we have seen that we put lone pairs of electrons into the equatorial positions to minimize electron pair repulsion. So we have 3 lone pairs of electrons that surround our central iodine. So we're going to put those 3 lone pairs equatorial. So we have our 3 lone pairs of electrons and equatorial like that, and then we still have 2 iodines. And so one iodine would have to go axial up here. And the same thing for the other iodine down here. And so I'll just leave off the lone pairs of electrons. And so this is what our ion looks like. Let's go back up here to our steps. And let's look at-- let's see, we've already done step 3. We predicted the geometry of the electron clouds to be trigonal bipyramidal. But when you're trying to predict the geometry of, in this case, the ion, you ignore any lone pairs of electrons. So let's go back down and look at it again. And so we're going to ignore the lone pairs of electrons around the central atom. And when we do that, we look for the shape. And we can see the shape is just a straight line, so we say it's a linear shape. So this is a linear shape, and since it's linear, we would predict the bond angle to be approximately 180 degrees since it's a straight line. And so that's the way to approach drawing this dot structure. And so we've done four examples of 5 electron clouds. And each example has been slightly different in terms of the number of lone pairs around the central atom.