So now we're going to look at the procedure for writing Lewis structures. We're going to talk about each of these steps and then we're actually going to go through a couple of examples to see how we would actually do this with an actual molecule. So the first thing we need to do, is to determine our total number of valence electrons, and that's what the examples we just looked at were about, is counting the number of valence electrons in each atom. And determining the total number of valence electrons in the molecule. Then we need to write our skeletal structure, and there's a few things we have to worry about with our skeletal structure. One, is the least electronegative atom occupies the central position. So we're going to have to remember our trend of going up and to the right, and fluorine being our greatest electronegativity of all the atoms. And we're going to determine the least electronegative atom. Now, sometimes hydrogen is not the least electronegative atom. But it's not going to be in the middle of the molecule, because hydrogen is unique in that it only can form one bond. And so hydrogen must be a terminal atom. And the same is true for fluorine. But these are always terminal atoms. So they are always on the outside of the molecule and never a central atom. Usually, when we look at the formula of a compound, what we see is that the least electronegative atom, or the central atom, is generally written first. So here we have NH3, regardless of whether nitrogen or hydrogen is the least electronegative. We have to remember that hydrogen must be a terminal atom. And we see that nitrogen is written first, and that's an indication that it's going in the central position. For sulphur, we know that it's less electronegative than oxygen, so it's also going into the central position. And this usually works, although there are some exceptions such as water, and oxygen is our central atom. But that's only going to happen when we're dealing with something like hydrogen. Okay. The other thing we're going to do is we're always going to connect everything with single bonds. Before we start dealing with any double bonds, we're going to try to make everything work with single bonds. And if all atoms can have their octet, have their eight electrons with just single bonds, we don't actually have to worry about forming those double bonds. So we're going to draw our skeletal structure, we're going to figure out how many electrons we've used in the bonds as we've drawn it. And we're going to determine the remaining electrons, so figure out what's left over to actually put into that structure. Because remember, the number of valence electrons tells us the number of electrons we have to find, places for in our final Lewis structure. Now we have our central atom, we have our terminal atoms, we have single bonds between them. Now we're going to take lone pairs of electrons and place them around the terminal atoms, those outer atoms, so that they each have an octet. Now hydrogen is kind of an exception here. Hydrogen only wants two electrons. And because it gets us two electrons by forming a single bond, we never put extra electrons around hydrogen. But if I had something like an oxygen, then I'm going to add electrons, as long as I have electrons, around that terminal atom until I've completed the octet. Now, this oxygen atom actually has eight electrons around it, six non-bonding and two bonding. If at that point we have additional electrons left over, then we place them around the central atom. So, anything that's left is going to go around that central atom, even if that means that we have more than eight, and we'll talk about some of those exceptions to this octet rule later in this unit. Now that we've distributed all the remaining electrons to the atoms, we need to check the number of electrons around each atom. And we're trying to identify here, which elements have an octet and which ones do not. And if we have atoms that do not have an octet, then we have to go to step number seven, which is moving one or more lone pairs of electrons from a terminal atom to form a multiple bond to a central atom, and as a result of that we get more electrons around the central atom. And we're going to continue to do that until we get an octet around every single atom. Important to remember, the total number of electrons used in this structure must equal the number of valence electrons. I cannot simply drop electrons somewhere on the way or find extras. The number of valence electrons, what we calculate in step one must match exactly the number of electrons in our Lewis structure at the end. Now, let's look at an example of how we can draw the Lewis structure for a molecule. We're going to look at water as an example. So we start by counting the number of valence electrons. Have hydrogen which has 1 electron and there are 2 of them so that gives me 2 valence electrons. Oxygen has 6 electrons, there's only one of that so that's six electrons. So I see, I have a total of eight valence electrons. So no matter what I do at the end, my Lewis structure must have eight electrons. Now, I want to draw my skeletal structure with single bonds. I know that the least electro-negative atom goes in the middle. However, hydrogen can never go in the middle, which means that oxygen must be in the middle of this molecule, because hydrogens must be terminal atoms. Now, I can figure out how many electrons I've used in bonds, I've used four, two in each bond. So now I have four electrons left, that I have to place into my Lewis structure. I'm going to place electrons on my terminal atoms, to fill their octet, or the duet for hydrogen. But I see that each hydrogen already has 2 electrons around it. So I don't need to add any of those electrons to the hydrogen. Then, I'm going to place the remaining electrons, I had 4 left, on my central atom, that oxygen. And that uses up all my electrons. I have four bonding electrons here, from the two single bonds between oxygen and hydrogen and four non-bonding from the electrons on the oxygen. So four bonding and four non-bonding. Now, I want to check that each atom has enough electrons around it. So I look at my hydrogen, and I see that hydrogen has two electrons, so hydrogen is fine. The other hydrogen is the same, it has two electrons, it doesn't want an octet, it only wants two, and then I look at my oxygen. And I see that I've got eight electrons around my oxygen. It doesn't matter that I've already counted those two electrons for the hydrogens. We can, because we're sharing electrons, we can basically double count them in that structure. So I see that every atom has enough electrons around it. And therefore, I don't need to look at my option to form double bonds for this particular molecule. Simply because, I've been able to satisfy the octet rule for every atom without forming double bonds. So let's look at an example, where we find out the lone pairs that are found in the Lewis structure of NH3. So when we look at the Lewis structure of NH3, what we see is that we have to have five electrons from the nitrogen, hydrogen we have three of those atoms times one electron and this also has eight electrons in it. I first draw my structure will all single bonds and that uses up six electrons so I have two electrons remaining. My hydrogens have their complete octets, so I don't need to put any electrons around those terminal hydrogen atoms and my last two electrons go here on the nitrogen. So now I have eight electrons in my structure, six bonding and two non-bonding. Now, some molecules are very easy, like water, NH3. We can look at all single bonds. We can satisfy the octet rule for all, all our atoms and everything is good. However, that's not always the case, we have some situations where we can't do it with just single bonds. And so we're going to look at an example here with CO2, we've actually already seen this earlier but we're going to look at how we got to that structure of CO2. So the first thing we're going to do, is we're going to count the total number of valence electrons. So we have carbon and oxygen. Carbon has 4 valence electrons and there is 1 carbon. Each oxygen has 6 valence electrons and there are 2 of them. And so we have 12 and 4 for a total of 16 valence electrons. So we know in our final structure, we can only use 16 electrons. So now I want to draw my skeletal structure, and I'm going to do this with all single bonds. There are two clues that tell me that the carbon is going to be the central atom. One, it's written first in the formula, and the electronegativity of carbon is less than that of oxygen. So I can draw my skeletal structure with my two single bonded oxygens. Now I want to see how many electrons I've used in bonds which is four. And I see that I've used up four electrons. So I have 12 electrons left to put into my structure. Now, I'm going to place electrons on my terminal atoms, so on my oxygen atoms until I've used them all up or until I've completed the octet on this terminal atoms, whichever one comes first. So here, I've done those at the same time. I've used up my 12 electrons. And completed the octet on both of those oxygen atoms. Now, we replace any remaining electrons on the central atom. However, I used all 12 of my electrons on the two oxygen atoms. So I don't have any electrons to put on the carbon atom. Now I want to check that each atom has satisfied the octet rule. Each oxygen has eight electrons around it, which is great. However, I notice that carbon does not. It only has four electrons. So I can't stop with step six on this problem, because I want to make sure that every atom as an octet of electrons. So the way that I'm going to do that, is I'm going to move electrons from my terminal atoms to form double bonds so that all atoms have eight electrons. I could also form mult, triple bonds if I need to. But in this case it's going to be double bonds. So here, I move a pair of electrons from the oxygen on the left. And when I do that, I still see that I have eight electrons around my oxygen on the left. And my carbon situation's improved, now I have six electrons around the carbon. But that's not where I want to be, so now I need to look at moving another pair of electrons. So I'm going to take a pair of electrons, and it doesn't really matter which ones it, they are, and I'm going to move them into that bond and make that single bond between carbon and oxygen a double bond. Now when I look at my structure, I see that oxygen has eight electrons on both sides, but carbon also has eight electrons because it has the four bonding groups where are each a pair of electrons. So we have a total of eight electrons around each atom. So now each atom is satisfying the octet role. So what we have here, is a summary of those seven rules that we look at when we're trying to draw Lewis structures. And the best way to learn how to draw Lewis structures is actually to practice it and to do it with a lot of different molecules and ions. So step one, determine valence electrons. Two, skeletal structure all single bonds. The most, the least electron negative atom going in the middle of the molecular unless, of course that's hydrogen which must be terminal. See how many electrons are left. Lone pairs on our terminal atoms. Any remaining lone pairs will go on our central atom. Then we have to check each atom and make multiple bonds if needed. So let's look at an example with HCN. Will all atoms in HCN have an octet if it contains only single bonds? So the answer is no, because when I look at hydrogen, I have 1 valence electron, carbon has 4 valence electrons, and nitrogen has 5 valence electrons, that gives me a total of 10 valance electrons to work with in my structure. If I draw the structure with my single bonds, because that's my simplest skeletal structure, I see that I've used four electrons in my two bonds, and I have six remaining. So I'm going to start by filling in those electrons on my terminal atoms. Hydrogen doesn't need any more electron it has its duet, it's perfectly fine. But I can fill in six electrons on the nitrogen. So now, nitrogen has an octet, hydrogen has its duet which is satisfying the octet rule, but carbon only has four electrons around it. How many bonds can hydrogen form? Right, just one. Hydrogen will only form one single bond to any other element and because of this this, limits some of our options of where we make double and triple bonds. We will never see a double or triple bond with hydrogen in any other atom. So, which of the following is the correct Lewis structure for HCN? So, now what we want to do is follow our rules. We already got to the point where we had our skeletal structure. And we showed our non-bonding groups on the nitrogen but I said my carbon was not happy because it only had four electrons. So now I need to look at moving some of these lone pairs of electrons from the nitrogen into a bond. Notice that hydrogen doesn't have any lone pairs of electrons, so we can't actually contribute anything to make a double bond and we said that hydrogen only forms single bonds. So if I form multiple bonds, it has to be between the carbon and nitrogen. Now there are some times where you don't clearly know where the double bond is going to form, and we'll actually talk about something called formal charge that helps us decide where that double bond or triple bond is going to go. But right now, we're just looking at simple examples where there's really only one option. So now, I'm going to take one of these pairs of electrons and move it in to form a double bond. When I do that, I see that helps the situation and I now have six electrons around carbon. But I need eight, so I have to do the same thing with another double bond, and form a triple bond. So, what I end up with, is a triple bond between carbon and nitrogen, and a lone pair of electrons on the nitrogen. Remember that I had ten valence electrons that I needed to make sure were in my final structure. So I see one pair. So that two electrons between hydrogen and carbon, six electrons, for a total of eight in my bonding groups and then I have a lone pair of electrons on the nitrogen, which gives me a total of ten electrons. So I know, that I have enough electrons in my structure. Let's look back at some of the examples and see what's wrong with these structures. Here, the problem is with carbon. It only has four electrons around it, and it wants eight. When I look at the second structure, I see that I have a double bond here between hydrogen and carbon, and I can actually form a double bond with hydrogen. When I look at number four, I see that nitrogen does not have an octet, and I also see that I put lone pairs of electrons on carbon before all of my terminal atoms had satisfied the octet rule. Same thing with number five, here I have nitrogen with only six electrons around it. But I also see a lone pair on the central atom that should've been added to the nitrogen before the carbon got any more electrons or got the lone pair of electrons, because nitrogen still has an incomplete octet. So this is our best structure. All of our atoms satisfy the octet rule, and we have ten valance electrons just as we should. So now, let's look at the Lewis structure for one of our ions. So in this example, we go through our number of electrons, which we determined earlier was 24 electrons, and now what we need to do is distribute those electrons, and we're going to do that first, through our single bonds. So, we've used up six electrons in the single bonds, so we have eighteen electrons remaining. Now, going to start adding those electrons to the terminal oxygen atoms. And I see that each of those needs six electrons to complete its octet. And so I have 18 electrons. Now, I've used all 24 electrons, six in bonding, 18 in non-bonding, and I have no more electrons left, but carbon does not have its octet. So, now what I need to do, is move one pair of electrons in to form a double bond. So I have oxygen, but that only has two pairs of non bonding electrons, because it moved two of them, the electrons to form a double bond with carbon. And I have still two single bonds between carbon and oxygen. And what I see know is that each oxygen has an octet. They each have eight electrons around them. My oxygen here with a double bond still has it's eight electrons, and carbon has eight electrons around it. Now we show the brackets around this just to indicate that it is an ion, remembering that the charge belongs to the whole ion, not to any one particular atom. Now, we showed the double bond here, what we're going to look at in the next unit is look at something called resonant structures, and we'll see why putting the double bond here versus putting it in one of the other two locations are all absolutely equivalent. The key is that we only need one double bond, because beyond that, making two double bonds as we did in this pro answer gives us an atom of carbon that now has 10 electrons around it, which violates the octet rule. So in the next unit, we're going to look at resonance structures and formal charges, helping us figure out what the best Lewis structure really is.
