 In this video, we're going to discuss the octet rule and Lewis structures for simple gist elements and for more complex compounds. If you recall from the periodic table, the noble gases, which are in group AA, are the least reactive elements on the periodic table. They are the least reactive because they have a full set of valence electrons. They have eight valence electrons. So all the other elements on the periodic table basically want to be like a noble gas and have eight valence electrons, and that's where the octet rule comes in. Atoms on the periodic table will gain, lose, or share valence electrons so that they achieve a full set of valence electrons, which is eight. Okay? However, there are a couple of exceptions to the octet rule, and those are hydrogen and helium. Once hydrogen and helium are on the first row of the periodic table, a full set of valence electrons consists of only two valence electrons. So hydrogen and helium will only have two. All the other elements on the periodic table will try to achieve eight valence electrons. We can represent the number of valence electrons in an element using Lewis symbols. We will use a dot to represent the number of valence electrons. Hydrogen only has one valence electron because it's in group 1A on the periodic table. Carbon has four valence electrons. It's in group 4A, so that group number when it's a representative element tells you how many valence electrons it has. Oxygen's in group 6A has six valence electrons, so we have six dots around the symbol for oxygen. And neon is a noble gas. It has eight valence electrons, so we have eight dots around it. Those are the Lewis symbols for just elements. We can make more complex Lewis structures for a whole compound that shows how the electrons are shared within that compound so that every element in the compound can achieve its octet or its full set of electrons. The steps for writing a Lewis structure are written on the board here. Basically there are three or four steps depending on the structure or on the compound. The first step is to sum up your valence electrons for all of the atoms in the compound. So you're going to go to the periodic table and you're going to see what group number each element is in and you're going to add up all the valence electrons. The second step will be to draw the skeleton structure which consists of basically putting the first atom that's in the chemical formula in the center and put the other atoms around the outside. There is an exception to that rule as well. Hydrogen can never be in the center. Once you have that drawn, you're going to connect all of your outer elements to the center atom with a line which represents a bond. Then we have to distribute our remaining electrons around so that everybody gets their full set of eight except for hydrogen which is only going to have two. At that point we'll check and make sure that everybody has their octet or their maximum number. If not, we go to step four and that's where we may have to make multiple bonds. So we're going to do a few examples and draw some Lewis structures for a few different compounds. The first compound we're going to do is methane, CH4, so here's our formula. First step, sum up the valence electrons. Carbons in group 4A, so it has four valence electrons. Hydrogen is in group 1A, it has one valence electron. There are four hydrogens in the compound so I'm going to multiply it by four, add them up, we have eight electrons. So we have eight electrons to distribute and share amongst all the atoms in that compound so that everybody has a full set of valence electrons. Step two, drawing the skeleton structure. The first atom in the formula goes in the center. The other atoms go around the outside and we're going to connect each of the outer atoms to the center with a bond. We use a line to represent a bond, a bond represents two electrons, a pair of electrons. So so far in our skeleton structure we have used two, four, six, eight electrons to connect the outer elements to the center element. We're going to subtract that from our total number of electrons and that tells us that we have used all of our electrons. So now we don't have any more to distribute so we're going to check and make sure that everybody has their full set of valence electrons. Hydrogen is an exception and only needs two. If we look at this hydrogen atom it has its two electrons because it's sharing these with carbon so it has two. This hydrogen has two, this hydrogen has two and this hydrogen has two so all the hydrogens are happy. Now we have to look at the center carbon. This carbon gets used of all of the bonded, the electrons in the bonds so it has two, four, six, eight electrons around it. That's its full set so this is the complete Lewis structure for methane. Let's do another example, phosphorus trichloride. First step, sum up the valence electrons. Phosphorus is in group 5A so it has five valence electrons. Chlorine is in group 7A so it has seven valence electrons. We have three chlorine atoms, multiply it by three. When we add it up we should have 26 electrons. So step two, our skeleton structure. This atom is going to go in the center. The remaining atoms go around the outside, connect each of them with the bond. The bond represents two electrons so we have used two, four, six electrons. I'm going to subtract it from our total number of electrons so we have 20 electrons left to share or to distribute. Third step, distribute the electrons, starting with the outer atoms. So let's pick a chlorine atom. This chlorine atom already has two electrons so it needs six more to complete its octet so we are going to add six more electrons. This chlorine already has two, we're going to add six more so that it completes its octet. Same for this, it has two, we add six more. So we have used six, twelve, eighteen electrons. We have two remaining, two electrons remaining. We have already given electrons to all the outer atoms to fulfill the octet. Any that are left over at this point will go to the center atom. So I'm going to give these two to phosphorus. In fact, we have used all of our electrons, we've distributed them all. So now we have to check, make sure everyone has the octet. If you look at this chlorine, it has all eight. This chlorine has all eight. This one has eight. Check the center atom. Phosphorus has two, four, six from bonds and the two that are a lone pair so it has eight as well. So this is the complete electron, or Lewis structure for phosphorus trichloride. We're going to do one more example that takes us down to step four. SO2, sulfur dioxide. First step, sum up your valence electrons. Sulfur is in group 6A so it has six. Oxygen's in group 6A. It has six and there are two of them. So we have 18 electrons to share. Draw our skeleton structure, S, two O's and you can put the O's anywhere you wish. Attach each with a bond. So we have used two, four electrons so far. So we have 14 electrons remaining. We need to distribute them starting with the outer atoms first. So this one needed six more to get eight. This oxygen needed six more to get eight so we've used 12. We have two left over. The left overs always go to the center atom. You could put it on top or bottom, doesn't matter. I've given those two to sulfur. We have no more electrons so we've distributed all the electrons. It's time to check. Does everybody have eight? This oxygen has all eight. This oxygen has all eight. Sulfur has two, four, six. Does not have all eight electrons. So we are not done. Sulfur has to have all eight. What happens at this point is that one of the outer atoms has to share more electrons. So the oxygen for example on either side doesn't matter which side you choose. They have all of these lone pairs that aren't being shared. So we can take one of these pairs and move it in between the S and the O so it's being shared. Now the sulfur gets to use those electrons along with the oxygen. So you have formed a double bond and the final structure would look like this. The lone pair that was here has moved in between. It's being shared. Now the sulfur has two, four, six, eight electrons and everyone is happy. The key here, the forming multiple bonds only happens if the center atom does not have its full octet after step three. So that is always a last resort to form multiple bonds. So with those three examples that we did I think you have a pretty good understanding of how to draw a Lewis structure for any type of covalent compound.