Class 11 Chemistry (India)
VSEPR for 2 electron clouds
The valence shell electron-pair repulsion (VSEPR) model is used to predict the shapes of molecules and polyatomic ions. VSEPR is based on the idea that the “groups” or “clouds” of electrons surrounding an atom will adopt an arrangement that minimizes the repulsions between them. In this video, we look at examples of molecules in which there are two groups of electrons around the central atom. Created by Jay.
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- At5:55, how does one know carbon does not have a formal dot charge of zero?(25 votes)
- Try calculating the formal charge. Remembering that the formula for formal charge is [# of valence electrons] – [electrons in lone pairs + 1/2 the number of bonding electrons]. So carbon has four valence electrons, right? Moving on to lone pairs, we can see that there are no lone pairs, so we can plug a 0 in for that. Finally, there are a total of 4 electrons in the bonds, but only half of these electrons go to the carbon (think about what a covalent bond is and this will make sense). So taking what I've said above, 4 valence - 0 lone pair electrons - (1/2)(4 bonded electrons) = 4-0-2 = 2+. Now, you could even look at the formal charge of the oxygens and it should make more sense as to why you take those pairs of electrons and put them into the double bonds (keeping in mind that 0 formal charge is the best on atoms). Hope this helped! ( I took the formula from http://www.masterorganicchemistry.com/2010/09/24/how-to-calculate-formal-charge/ and I think it's a good link to look at!)(39 votes)
- What do you do if there ARE lone pairs on the center atom?(17 votes)
- The short answer is that the initial configuration is determined by how many electron regions exist on the central atom. Then, the molecular geometry is determined by considering how many actual atoms are in the structure, as opposed to electron pairs.
For example, if you are considering water (H20) which has a central Oxygen, with two pairs of electrons, bonded to two Hydrogen atoms.
The central Oxygen would have 4 regions, one region for each of the electron pairs, and one region for each of the single bonds to hydrogen. As a result, it's initial geometry would be tetrahedral.
Now, not all of those 4 regions are filled with an atom so we have to think about what happens if only half of those regions are filled. (The video on VSEPR 4 can better illustrate this.) What you want to keep in mind is that the atoms want to be as far apart from each other as possible, and the electron pairs repel atoms a little more then atoms repel atoms. With this in mind, the tetrahedral is drawn with atoms on one side, electrons on the other. The final molecular geometry ends up being bent.(22 votes)
- Going by this, the H2O molecule would seem to qualify for 2 electron clouds and linear geometry with a bond angle of 180 degrees. Why is it instead considered "bent" geometry with a bond angle of 104.5 degrees?
Apologies if there is a video that addresses this and I haven't seen it yet.(13 votes)
- The point is that the oxygen atom has two non-binding electron pairs as well. They count as electron clouds.
These non-binding pairs will repel the hydrogen atoms slightly.
If you still did not quite understand this, watch the VSEPR for 4 electron clouds video. This exact example is addressed at the end of it.(7 votes)
- Isn't Be has a full octet?
I see only two bonds connecting Be and two chlorine atoms.
I thought it should be double bond and there might be two lone pairs for each of chlorine atom .(8 votes)
- Be does not obey the octet rule: Be seeks 4, not 8 electrons. Thus, Be (unlike the other members of Group 2) makes only covalent bonds.
While there are other elements that do not follow the octet rule (H, He, Li and usually B), Be is rather unique in many of its properties.(17 votes)
- At1:56how come we don't put 2 more lone pairs of electrons on Be?(3 votes)
- Total number of electrons in BeCL2 = 16
Which are all used up. ( 6 each to Cl + 4 electrons as bonds)(5 votes)
- At0:38he says that you can find Beryllium in group two on the periodic table. None of the tables I've looked at have group numbers. How can if find the groups->number of valence electrons on a periodic table?(3 votes)
- Almost every periodic table I can see on Google image search has the group numbers on them.
Even if there aren't group numbers you could write them in, they go from 1-18. The left most group is 1, the right most is 18.(2 votes)
- why does carbon not have a formal charge of 0(2 votes)
- Formal charge = valence electrons - lone pair electrons - bonds
Using this formula, carbon has a formal charge of 0 when it has 0 lone pairs and 4 bonds.
4 - 0 - 4 = 0
If carbon dioxide looked like O-C-O with 3 lone pairs on each oxygen, then carbon only has 2 bonds so it has a formal charge of +2 (4 - 0 - 2 = +2) and each oxygen has a formal charge of -1 (6 - 6 - 1 = -1)
In the structure with 2 double bonds (like O=C=O) all atoms have a formal charge of 0.(3 votes)
- What is the difference between electron domain geometry and molecular geometry?(1 vote)
- The electron domain geometry includes the geometry of both lone pair electrons and bonds (i.e. all electron domains). The molecular geometry tells the shape that only the bonds make (i.e. any position with a lone pair isn't part of the shape in molecular geometry).(4 votes)
- hi there, my question is does a radical count as 1 electron cloud? For example, is NO2 bent (if a radical counts) or linear (if a radical does NOT count). Also, is the nitrogen in NO2 sp2 or sp hybridized? Thanks in advance!(2 votes)
- The radical electron should still be counted as an electron cloud as NO2 is has bent molecular geometry
The nitrogen atom would be considered sp2 hybridised.(2 votes)
- At5:42, doesn't the structure tell us that each oxygen atom has 7 valence electrons?(2 votes)
- I can see what you are saying and why you may think that. At that point in the video, carbon is shown to have donated all four of its valence electrons to form the two single bonds with oxygen. This would give carbon a formal charge of +2 and each oxygen a formal charge of -1. This formal charge of -1 on each oxygen arises from an extra electron on each oxygen, which is in addition to its 6 valence electrons. This gives the impression of each oxygen having 7 valence electrons but this is only because each has an extra electron giving it a net negative charge.
As the video moves on, so oxygen donates electrons to make double bonds. In the final structure, all formal charges are zero and each oxygen now has two lone pairs and has contributed two of the electrons to the double bonds. Hence, by the time the structure has been completely drawn, oxygen is back to having 6 valence electrons.(2 votes)
This next set of videos, we're going to predict the shapes of molecules and ions by using VSEPR, which is an acronym for valence shell electron pair repulsion. And really all this means is that electrons, being negatively charged, will repel each other. Like charges repel, and so when those electrons around a central atom repel each other, they're going to force the molecule or ion into a particular shape. And so the first step for predicting the shape of a molecule or ion is to draw the dot structure to show your valence electrons. And so let's go ahead and draw a dot structure for BeCl2. So you find beryllium on the periodic table. It's in group 2, so two valence electrons. Chlorine is in group 7, and we have two of them. So 2 times 7 is 14. And 14 plus 2 gives us a total of 16 valence electrons that we need to account for in our dot structure. So you put the less electronegative atom in the center. So beryllium goes in the center. We know it is surrounded by two chlorines, so we show beryllium bonded to two chlorines here. And we just represented four valence electrons. So here's two valence electrons. And here's another two for a total of four. So, instead of 16, we just showed four. So now we're down to 12 valence electrons that we need to account for. So 16 minus 4 is 12. We're going to put those left over electrons on our terminal atoms, which are our chlorines. And chlorine is going to follow the octet rule. Each chlorine is already surrounded by two valence electrons, so each chlorine needs six more. So go ahead and put six more valence electrons on each chlorine. And, since I just represented 12 more electrons there, now we're down to 0 valence electron. So this dot structure has all of our electrons in it. And some of you might think, well, why don't you keep going? Why don't you show some of those lone pairs of electrons in chlorine moving in to share them with the beryllium to give it an octet of electrons? And the reason you don't is because of formal charge. So let's go ahead and assign a formal charge to the central beryllium atom here. So remember each of our covalent bonds consists of two electrons. So I go ahead and put that in. And if I want to find formal charge, I first think about the number of the valence electrons in the free atom. And that would be two, four-- four berylliums. So we have two electrons in the free atom. And then we think about the bonded atom here, so when I look at the covalent bond, I give one of those electrons to chlorine and one of those electrons to beryllium. And I did the same thing for this bond over here, and so you can see that it is surrounded by two valence electrons. 2 minus 2 gives us a formal charge of 0. And so that's one way to think about why you would stop here for the dot structure. So it has only two valence electrons, so even though it's in period 2, it doesn't necessarily have to follow the octet rule. It just has to have less than eight electrons. And so, again, formal charge helps you understand why you can stop here for your dot structure. Let me go ahead and redraw our molecule so we can see it a little bit better. And we'll go ahead and move on to the next step. So let me go ahead and put in my lone pairs of electrons around my chlorine here. So we have our dot structure. Next, we're going to count the number of electron clouds that surround the central atom. And I like to use the term electron cloud. You'll see many different terms for this in different textbooks. You'll see charge clouds, electron groups, electron domains, and they have slightly different definitions depending on which textbook you look in. And really the term of electron cloud helps describe the idea of valence electrons in bonds and in lone pairs of electrons occupying these electron clouds. And you could think about them as regions of electron density. And, since electrons repel each other, those regions of electron density, those clouds, want to be as far apart from each other as they possibly can. And so let's go ahead and analyze our molecule here. So surrounding the central atom. So we can see that here are some bonding electrons right here surrounding our central atom. So we could think about those as being an electron cloud. And then over here we have another electron cloud. So we have two electron clouds for this molecule, and those electron clouds are furthest apart when they point in opposite directions. And so the geometry or the shape of the electron clouds around the central atom, if they're pointing in opposite directions, it's going to give you a linear shape here. So this molecule is actually linear because we don't have any lone pairs to worry about here. So we're going to go ahead and predict the geometry of the molecule as being linear. And if that's linear, then we can say the bond angle-- so the angle between the chlorine, the beryllium, and the other chlorine-- is 180 degrees. So just a straight line. All right. So that's how to use VSEPR to predict the shape. Let's do another example. So CO2-- so carbon dioxide. So we start off by drawing the dot structure for CO2. Carbon has four valence electrons. Oxygen has six. And we have two of them. So 6 times 2 gives us 12. 12 plus 4 gives us 16 valence electrons to deal with for our dot structure. The less electronegative atom goes in the center, so carbon is bonded to oxygen, so two oxygens like that. We just represented four valence electrons. Right? So two here and two here, so that's four. So 16 minus 4 gives us 12 valence electrons left. Those electrons are going to go on our terminal atoms, which are oxygens if we are going to follow the octet rule. So each oxygen is surrounded by two electrons. So, therefore, each oxygen needs six more valence electrons. I'll go ahead and put in six more valence electrons on our oxygen. Now you might think we're done, but, of course, we're not because carbon is going to follow the octet rule. Carbon does not have a formal charge of 0 in this dot structure, so even though we've represented all of our valence electrons now, we need to give carbon an octet. We need to give carbon a formal charge of 0. And we can do that by moving in this lone pair of electrons into here to share those electrons between the carbon and the oxygen, and also with this lone pair of electrons. So we move those in like that. And now we can see that carbon is double bonded to our oxygen. So now our dot structure looks like this. And each oxygen, instead of having three lone pairs of electrons, now each oxygen only has two lone pairs like that. So there is our dot structure. So let's go back up here to look at our steps for predicting the shape of this molecule. So step 1 is done, draw dot structure to show the valence electrons. Next, we're going to count the number of electron clouds surrounding our central atom. So we go back down here, and we find our central atom, which is our carbon. And we think about the regions of electron density that surround that. So we can count this double bond as a region of electron density because we're not worried about how many electrons are there. We're just worried about the fact that there is a region of electron density. So that's one electron cloud. And then over here we have another electron cloud. So we have two regions of electron density. We have two electron clouds here, which are going to repel each other. So when we look at step 3-- predict the geometry of the electron clouds-- predict the geometry of the electron clouds around the central atom. Well, those electron clouds are going to be opposite to each other. They're going to point in opposite directions. So, once again, they're going to force this molecule into a linear shape. So this carbon dioxide molecule is also linear with 180 degree bond angle. So, once again, we don't have any lone pairs of electrons on our central atom, so we don't really have to worry about that. And we can go ahead and predict the geometry as being linear. So that's how to approach it. Draw the dot structure. Think about electron clouds and think about the shapes of your molecules. In the next video, we will look at how to approach three electron clouds.