- Drawing dot structures
- Drawing Lewis diagrams
- Worked example: Lewis diagram of formaldehyde (CH₂O)
- Worked example: Lewis diagram of the cyanide ion (CN⁻)
- Worked example: Lewis diagram of xenon difluoride (XeF₂)
- Exceptions to the octet rule
- Counting valence electrons
- Lewis diagrams
- Resonance and dot structures
- Formal charge
- Formal charge and dot structures
- Worked example: Using formal charges to evaluate nonequivalent resonance structures
- Resonance and formal charge
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- More on the dot structure for sulfur dioxide
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
- Molecular polarity
- 2015 AP Chemistry free response 2d and e
Guidelines for drawing Lewis dot structures. Created by Jay.
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- Why does the least electronegative atom belong in the the center? Doesn't the central atom receive electrons from other atoms, so wouldn't it make sense for the most electronegative one to go in the center?(20 votes)
- no because in order to satisfy the octet rule which says the second and the third period elements under the neither electropositive nor electronegative must not become the non-central atom in the compound. oxygen satisfies the octet rule as the most electronegative and carbon the least electronegative between the two and hence we see the electron dot structure as this.(4 votes)
- Does the central atom always form a double bond with some other atom to complete its octet, or can it also have a lone pair instead?(7 votes)
- There are few rules in chemistry that are "always" followed by all substances.
Sometimes central atoms have double bonds to complete the octet, sometimes lone pairs, and sometimes they just plain don't follow the octet rule (the octet rule is not a firm law of chemistry: though it is very often followed there are numerous exceptions).(17 votes)
- This is killing me right now, one of the MANY things i'm confused about..... wouldn't an oxygen bond be left with 4 electrons, not 6 because it's already sharing two? And how do you know when the structure is bent or not?(8 votes)
- You can tell a structure is bent if there are lone pairs left behind. Oxygen and Ammonia are bent because they have lone pairs.
An oxygen normally has 6 electrons in it's outer shell (aka valance electrons because they're on the outermost shell). It wants 8 electrons (octet rule). It gets 2 more by sharing it with other atoms, like 2 Hydrogens.(5 votes)
- At13:40, the description of the molecule/cation shows that Xe has five co-valent bonds plus two additionally allocated electrons... my question is where do these 12 electrons "reside" 5s and 5p together give six quantum combinations. The 4d is already populated and was not counted when totaling up the number of valence electrons. Do the bonds invoke or consume some of the 6s and 5d orbitals?(4 votes)
- You may not yet have gotten to these topics. One video is "VSEPR for 6 Electron Clouds". Then you look at the videos on "Hybridization and Atomic Orbitals".
The Xe atom says to itself, "I need six orbitals to hold 12 electrons. I can use the 5s and the three 5p orbitals, but I will have to use two of my empty 5d orbitals. I will temporarily take out the electrons from the 5s and 5p orbitals. Then I will hybridize all six empty orbitals to form six new equivalent sp³d² orbitals."
"Now I have the 8 electrons from the 5s and 5p orbitals, plus 1 from each of the 5 F atoms, minus 1 that I lost to form a cation. That makes 12 electrons, so I have just enough sp³d² orbitals to hold all the electrons."(13 votes)
- Could you please explain how you decide--which rule or criteria you use--which of the two atoms is "least electronegative"?(2 votes)
- It is one of the trends of the periodic table. Electronegativity increases upwards and from left to right. So, fluorine would be the most electronegative element.(15 votes)
- How come the central atom can exceed an octet?(3 votes)
- For elements from phosphorus up, the d orbitals are close in energy to the s and p orbitals, so they can use one or two d orbitals as well to form bonds.(4 votes)
- At5:28, how do you know that it can reach octet? Is there a proof or a way to determine? Does it have 8 valence electrons just because it is in the second period? Then what about other periods?(3 votes)
- In NO2+ how come N has 4 bonds? I thought N could only have a maximum no. Of 3bonds??(2 votes)
- Nitrogen will always obey the octet rule. Nitrogen will only have three bonds if there is a lone pair attached to the nitrogen as in NH3 (ammonia).
NH4+ (ammonium) is another very common example of 4 bonds to Nitrogen. If there are four bonds to the nitrogen, the nitrogen will always carry a + charge.(4 votes)
- So what would be the Lewis Dot Structure for the compound of N2O2? I understand the valence electron part of it, but I don't understand the actual drawing of the Lewis Structure.(2 votes)
- http://www.chemspider.com/Chemical-Structure.3874228.html This shows the bonds of the H2O2.
https://images.search.yahoo.com/images/view;_ylt=AwrB8pg3sB9XM0UA2CgunIlQ;_ylu=X3oDMTIycnRndjAyBHNlYwNzcgRzbGsDaW1nBG9pZAMyZGI3ZWI3MWY0ZDAzMmM1MmExN2ZlODQzYmI0MDc5YQRncG9zAzIEaXQDYmluZw--?.origin=&back=https%3A%2F%2Fimages.search.yahoo.com%2Fyhs%2Fsearch%3Fp%3DN2O2%26fr%3Dyhs-mozilla-002%26ri%3D7%26hsimp%3Dyhs-002%26hspart%3Dmozilla%26tab%3Dorganic%26ri%3D2&w=486&h=1020&imgurl=2.bp.blogspot.com%2F-pMqQAsq9P8M%2FT0Fgt6rIgaI%2FAAAAAAAAAKM%2FVGP9wQS0aA4%2Fs1600%2Fn2o2_final.png&rurl=http%3A%2F%2Fimgarcade.com%2F1%2Fn2o2-lewis-structure%2F&size=51.8KB&name=%3Cb%3EN2o2%3C%2Fb%3E+Lewis+Structure+Figure+1%3A+lewis+structures+for&p=N2O2&oid=2db7eb71f4d032c52a17fe843bb4079a&fr2=&fr=yhs-mozilla-002&tt=%3Cb%3EN2o2%3C%2Fb%3E+Lewis+Structure+Figure+1%3A+lewis+structures+for&b=0&ni=160&no=2&ts=&tab=organic&sigr=11ccfb4ap&sigb=13n6qctib&sigi=12novehts&sigt=11q5qb3do&sign=11q5qb3do&.crumb=MgZEsw/1Qpl&fr=yhs-mozilla-002&hsimp=yhs-002&hspart=mozilla This shows the Lewis Structure.
I noticed that the first link has double bonds between the N and the second link has triple bonds between the N.(2 votes)
Here's some of the guidelines for drawing dot structures. So let's say we wanted to draw the dot structure for this molecule, so silicon tetrafluoride. The first thing we would need to do is to find the total number of valence electrons. And we would account for these valence electrons in our dot structure. So to find the valence electrons, we need to look at it periodic table. So here I have a modified version of the periodic table. And you see what I've done is, I've kind of cut out the d-block here, so we can focus in on the elements that we're going to be drawing in our dot structures. It also makes it easier to see how the group numbers correspond to the number of valence electrons. For example, if we look at elements in the first group, like hydrogen, lithium, or sodium, the first group all have one valence electron. So the group number corresponds to how many valence electrons something has. So hydrogen has one valence electron. If we think about the periods, hydrogen is in the first period, or the first energy level. So periods are going horizontally across your periodic table. Hydrogen is in group one, it has one valence electron. And if I go over here to helium, this would be two valence electrons for helium. And so in the first energy level, you can fit a maximum of two electrons. And so this is going to important when we're drawing our dot structures. Because when we're drawing hydrogen, we're always going to surround it by two electrons, or a single covalent bond. When you take the second period on the periodic table, so here we are on the second period, lithium has one valence electron, beryllium has two, boron has three, carbon has four, nitrogen has five, oxygen has six, fluorine seven, and neon eight. And so you have more orbitals in a second energy level. And so because of that, you can fit more electrons. So maximum of eight electrons in the second energy level. And this is where the idea of the octet rule comes in. So for elements like carbon, nitrogen, oxygen, fluorine, understanding the octet rule is going to help you when you're drawing dot structures. Now it is possible for some of the elements of the second period to not have eight electrons. It's possible for them to have less than eight electrons. So things like boron will sometimes do that. But it is not possible for elements to have more than eight electrons. Always check your dot structures, and make sure that if you have an element in the second period, you do not exceed eight electrons. Once you get to the third period, you have even more orbitals available to you now. So in the first energy level, you have only one s orbital. In the second energy level you have s and p orbitals, and in the third energy level you have s, p, and d orbitals. So you can fit more than eight electrons. And so therefore it's possible to exceed the octet rule for elements in the third period and beyond. And we will see a few examples of that in this video, and some of the ones to come here. So getting back to our molecule, silicon tetrafluoride, if I wanted to find out how many total valence electrons are in this molecule, I need to find these elements on my periodic table. So I go over here and I find silicon, and I see it's in group four. So therefore one atom of silicon has four valence electrons. Fluorine is over here in a group of seven, and so therefore each atom of fluorine will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot structure, since it is the least electronegative of those two. So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover electrons to the terminal atoms. So in this case, the terminal atoms would be the fluorines. So let's go back down here and look at our dot structure. So fluorine would be the terminal atoms, and we're going to assign electrons to those fluorines. But how many do we need to assign? Well, going back to our periodic table over here, so fluorine is in the second period. So pretty good bet it's going to follow the octet rule here. So we need to surround each fluorine atom with eight electrons. Each fluorine already has two electrons around it, so I'm going to go ahead and put six more around each fluorine, like that. So each fluorine get six more valence electrons. And since I'm assigning six valence electrons to four fluorines, 6 times 4 gives me 24. And so therefore we've now accounted for all of the valence electrons. And so this should be the should be the final dot structure here. And so we don't even need to go on to step four for this molecule. This is a very simple molecule to draw. Let's go ahead and look at another example. So now we have CH2O, which is the dot structure for, which is the molecular formula, I should say, for formaldehyde. And so if we're following our guidelines here, the first thing we need to do is find out the valence electrons. So we need to look and find carbon, hydrogen, oxygen on our periodic table. So let's go back up here, and let's find those elements here. So here's carbon. Carbon's in group four, so therefore carbon will have four valence electrons. Here's hydrogen, in group one, so one valence electron. And oxygen is in group six, so six valence electrons for oxygen. So let's go back down and let's calculate the total number of valence electrons that we need to represent in our dot structure. We have one atom of carbon, so that's four valence electrons. Each hydrogen is one valence electron, but we have two of them, so 1 times 2. And each oxygen is six, and we have one of them, so six. So 6 plus 4 plus 2 is 12 valence electrons. So we need to represent 12 valence electrons in our dot structure. All right. Let's go back up to our guidelines and see where we are now. So we've already figured out the total number of valence electrons. Once again, we don't have any charges, so we don't need to worry about the rest of step one. Step two is to assign the central atom, which is the least electronegative. And remember you're going to ignore hydrogen for this example. So if we ignore hydrogen, the central atom is either going to be carbon or oxygen. And let's look at our periodic table to figure out the relative values for those. Oxygen is more electronegative than carbon. Here we go. Here's oxygen, right next to fluorine. And here's carbon, over here. So remember your trends for electronegativity. Oxygen is more electronegative, therefore carbon is going to be at the center of our dot structure. So we're going to go ahead and put carbon in the center. So go ahead and draw carbon. And we know that the carbon's going to be bonded to two hydrogens. And we know that the carbon is going to be bonded to an oxygen here. So let's see how many valence electrons we've accounted for so far. Two, four, and six. So we needed 12. We just used up 6. So right now we have six valence electrons leftover. OK, so let's go back up and look at the next step here. So we are to step three. Assign the left over electrons to the terminal atoms here. All right, so the terminal atoms. Well, in this case, our terminal atoms would be hydrogen and oxygen. But we're not going to assign any electrons to hydrogen, because each hydrogen is now surrounded by two electrons. And so therefore we're going to assign those leftover electrons to the oxygen here. And oxygen's going to follow the octet rule. So it already has two around it, so it needs six more. So 2, 4, and then 6. And so that actually takes care of all of our valence electrons, right? So now we have accounted for all 12 of them. However we're not done with our dot structure here. So carbon actually follows the octet rule almost all of the time. And let's go back up here and let's look at step four here. So for step four, if the central atom doesn't have an octet, and it usually does have an octet, you can give it an octet by creating multiple bonds. So let's look at what we have so far for our dot structure, and let's see if we can create multiple bonds somehow. So if I took one of these electron pairs here, and I moved them into here, let's see what that would give us for a dot structure. Now I'd have carbon with a double bond to oxygen. And this oxygen would have two lone pairs of electrons. And then these hydrogens right down here. And so this would be the correct dot structure for formaldehyde. You have an octet of electrons around carbon. You have an octet of electrons around oxygen. And hydrogen has two electrons around it, which it is happy with. So this is the correct dot structure. And let me just talk about some terminology really fast here. So these electrons in here, so this would be a double bond between the carbon and the oxygen, these electrons you would call bonding electrons. Because, obviously, they're involved in bonding. And then these electrons out here would be non-bonding electrons, or lone pairs of electrons. So that's just terminology that you'll hear people use when you're talking about dot structures. All right. And then let me just go ahead and highlight the fact that carbon has an octet of electrons here. So this would be two, four, six, and eight electrons surrounding our carbon. All right. Let's do one more example here. Let's look at, this time an ion. So this would be the xenon pentafluoride cation here. So we need to look and find xenon and fluorine on our periodic table. So we can figure out how many valence electrons we're dealing with here. So let's find xenon first. So here is xenon. It is in group eight, so eight valence electrons for xenon. And then we've seen fluorine of course in group seven. So let's go back down and figure out how many valence electrons we need to account for here. So we have one xenon, and that's eight valence electrons. Each fluorine is seven, and we have five of them. So 7 times 5 is 35. So we have 35 plus 8 gives us 43. Now this is actually not the total number of electrons that we're going to account for, because this is an ion. This is a plus 1, a positive charge. Which means that we lost an electron. Remember, and electron is negatively charged, so if you lose a negative charge, you have a positive charge left over. And so we're going to take away an electron. So instead of 43, we are now going to account for 42 in our dot structure. Let's go back up to our guidelines, and I'll show you where I have that written down right up here. So find the total number of valence electrons, and then, in this case we had a positive charge in or ion. So we subtracted an electron. And then, again, step two, find the central atom, least electronegative. OK. So we know that fluorine is the most electronegative, so that means we're going to put xenon in the center. So we'll go ahead, and let's get some room right down here, and we'll go ahead and put xenon in the center, like that. On a two five fluorine, so we can go ahead and put in these bonds around here, like that. And how many valence electrons have we accounted for so far? Well, let's see. That would be two, four, six, eight, and ten. So we had to represent 42, and we just represented ten. So now we have 32 left over that we need to represent. And notice, xenon is already violating the octet rule. It's exceeding the octet rule. And it's OK. It can expand its outer shell, because xenon is past the third period on the periodic table, obviously, like what we talked about earlier. So we can go ahead and go to the next step now. We're going to assign the leftover electrons to the terminal atoms. And our terminal atoms are of course the fluorines, which we know are going to follow the octet rule. And so, once again, each fluorine has two electrons around it. So that means I'm going to give each of the other fluorines six more. So I'm going to go ahead and put six more valence electrons on each of my fluorines. And so when I'm thinking about how many valence electrons I have represented now, right, so I have six more on five fluorines. So 6 times 5 is 30. So I've represented 30 more electrons. And so that means I have two left over. So two valence electrons leftover. We haven't had this situation before. Let's go back up to the guidelines to refresh our memory, what we do, when we have leftover electrons after step three. So we get to step four here. So, if necessary, in this case it is, we're going to assign any leftover electrons back to the central atom this time. And if the central atom has an octet or exceeds an octet, which is what we have in this example, you are usually done. And we're not going to worry about formal charges in this video, I'll talk about them in the next video. So we're going to take those two electrons and assign them into our xenon. So let's go back down here, and assign those two extra electrons to the xenon. And so we would draw it like that. And so now we are done, we've represented all 42 of the valence electrons that we were supposed to. So you can count all those up if you want to. Now most people will represent this dot structure by putting brackets here. And putting a positive charge outside of it. So there's your xenon pentafluoride cation. So we're going to a lot more examples for drawing dot structures in the next several videos, and see how drawing dot structures allows you to predict the shapes of different molecules.