In the previous lecture we made enormous progress in the development of our Lewis Structure model of how molecules form and what determines the types of molecules which will form. Remember, that our Lewis Structure model is based on the octet rule. An experimentally derived rule that says that atoms typically bond so as to complete an octet of valence electrons, 8 electrons in their valence shell. And that they do so by sharing electrons with other atoms in pairs. We call a shared pair of electrons a covalent bond. We also learned that carbon atoms in particular are able to form single bonds by sharing a single pair of electrons, double bonds by forming, or by sharing two pairs of electrons and triple bonds by sharing three pairs of electrons. We need to extend though, beyond just doing carbon atoms and hydrogen atoms, which is all we did in the previous lecture. Let's try fluorine for example. Fluorine atoms we know, have seven valence electrons, and the valence is equal to 1. There's the experimentally derived octet rule, 7 plus 1 being 8. What will our Louis structure look like for fluorine? Well, let's see. It's got seven valance electrons, and we need to draw a structure that accurately represents the fact that it has a valance of 1. If we were to simply draw the Lewis structure to have 7 electrons sort of widely distributed around the fluorine, that would suggest that there would be seven unpaired, unshared electrons. And we might think that the valance of fluorine was 7, it's not, it's only 1. So this is not an appropriate structure. Instead, since we know that paired electrons typically are already going to not form bonds, then and we only want 1 unshared, unpaired electrons. I'm going to draw a flourine to look like this, because this gives both a single vacancy a single place where we can form a bond and the other electrons here are already paired and therefore probably not going to form bonds. We can use this model to account for example for the stability hydrogen fluoride molecule because if we then take this fluorine, which we have drawn, as so. And bring in a hydrogen with its single electron. Then we wind up with HF with a single bond in here as we would have expected. It also accounts for the stability of the F to molecule remember that fluorine is 1 of the diatomic gases that we studied early on. This suggests that we take the fluorine drawn as follows and a second fluorine and let them share there unshared, unpaired, electrons from before. And again form a single bond, and in fact the experimental data for Fluorine F2 is consistent with this bond strengthened, bond length being a single bond. So now we have a model for Fluorine. Let's try Oxygen. Oxygen you recall has 6 valence electrons, we derived that from our ionization energy data, and it has a valence of 2 such as in water. There's the octet rule again. 6 plus 2 is equal to 8. How shall we draw Oxygen? Based upon the lessons learned from flourine we're going to draw oxygen with 2 unshared, unpaired electrons and the remaining 4 electrons existing in pairs. And that gives us the opportunity for example if we want to draw water to draw it by taking the two unshared paired electrons, unpaired electrons. and pairing them up with an electron from each of the 2 hydrogens to form single bonds to each hydrogen, with 2 long pairs on the oxygens. Similarly, it helps us understand the strength of the bond in the oxygen diatomic molecule, oxygen was also one of the diatomic gases we learned about early on. So we'll take the two oxygens and now if I were to take the remaining electrons on either side I'd probably will want to share those so that I can complete an octet of both, for both oxygen atoms and correspondinly then we'll drawn oxygen with a double bond, and each oxygen having 2 lone pairs on it. And again, the data for oxygen suggests that oxygen is doubly bonded. How about nitrogen? Turns out the bond in nitrogen is even stronger than the bond in O2. So we will then consider what the nitrogen looks like, again based upon our lessons from oxygen, and from fluorine. Were not going to draw this with five electrons scattered around here, rather we will pair 2 of the electrons, and leave 3 of the electrons unpaired. That allows us to account for example,fFor the valance of 3 of a nitrogen atom and for example, the molecule ammonia will take the nitrogen, one pair of electrons, 3 unshared, unpaired electrons. We then share those electrons with each of 3 hydrogen atoms resulting in 3 single bonds, in the NH3 molecule. The nitrogen now has completed its octet. Furthermore, we can look at the M2 diatomic molecule and if we now simply take each nitrogen to have 1 lone pair on it and share the remaining electrons, we wind up with something that looks like a triple bond between the nitrogens in much the way we had triple bonds in the carbon atoms and in fact, the experimental data is consistent with the into bond being an extremely strong bond consistent with a triple bond. Well what we can now do is extend molecules in which we combined all the different types of atoms together and we can begin to develop a systematic means of drawing these molecular structures. And here's how our systematic method is actually going to work. Let's consider a particular molecule CH3NH2, methylamine. This is actually a stable molecule. It's an important molecule as well, in all kinds of syntheses. The first thing that we want to do is actually figure out how many available electrons are there in this molecule to be shared or to be alone in each of the atoms. To do that what we need to do is count up the number of valence electrons in the carbon atoms, the number of valence electrons in the nitrogen atom, plus one each for the hydrogen atoms. So the number is 4 plus that's for the carbon, plus 5 for the nitrogen, plus 5 times one, where there's one from each of the hydrogens, 4 plus 5 is 9 plus 5 more gives us a total of 14 available electrons. This is an extremely important calculation, to do, because at the end of our determination to molecular structure, the number of electrons that we have drawn had better actually be 14 or we've made a mistake. How many electrons do we need for every atom from the main group to have a, an octet of electrons or for each hydrogen to have 2? The number that we need is equal to, let's see, carbon needs 8, nitrogen needs 8 each hydrogen needs 2. That's a total of 26. So we need 26 electrons, to satisfy the valence, of each of the atoms in Methalamine. But there's no way we've got 26, we've only got 14. So how can we possibly do this? The answer is, we've got to share electrons. How many do we have to share? That turns out to be an easy calculation. It's the number we need minus the number that we have. So for example, in this particular case is 26 minus 14, we're going to share a total of 12 electrons. How are we going to share those 12 electrons? The answer is we're going to calculate the number of bonds. The number of bonds needs to be half of the number of shared electrons since we share electrons in pairs so, it needs to be 6. Now that we know that there are 6 bonds present, let's see if we can draw them. Well let's see,carbon is clearly going to be bonded to nitrogen can't can't form this molecule without that. And then, beyond that, there are going to be 5 hydrogens. We know carbon has a valence of 4, so it probably wants3 hydrogens around it. Nitrogen has a valence of 3, so it probably wants 2 hydrogens around it. At this point it would appear that we've actually completed our Lewis Structure, so lets check and see. We have2, 4, 6, 8, 10, 12 electrons remember each of these bonds is a shared pair of electrons, but wait that's only 12, the number that were available were 14, there has to be 2 more electrons. Furthermore, nitrogen has not satisfied the octet rule. It has only 6 electrons around it. We can solve this problem by saying the two extra electrons here that are not yet shown become a lone pair on the nitrogen atom. This systematic process that we've drawn on this particular slide is a good way to solve problems for an enormous number of molecules, in fact, it works rather well in general. Let's do another example. This particular example is going to simply be called formaldehyde. It is CH2O. And let's see, the number of available electrons is going to be equal to4 for the carbon, 6 for the oxygen, and 2 for each hydrogen. So there are 12 available. The number we need is 8 for the carbon, 8 for the oxygen, and we need 2 each for the 2 hydrogens, so that's 20 all together. Since we don't have enough electrons to satisfy the octet rule forever, we're going to have to share. The number we will share is 20 minus 12 is equal to 8 and correspondingly there should be 4 bonds in this molecule. Let's draw a picture now. Carbon and the oxygen will be bonded together. There are a variety of different ways we can try this but one of the ways that we'll try it initially is to simply say that the hydrogen will have both of the I'm sorry, the carbon will have both of the hydrogens on it here. Let's see, we used up all the electrons by forming these bonds, 1, 2, 3, that's 6 electrons, there are actually 12 around. Furthermore, we're suppose to have four bonds. We only have three bonds. A fourth bond actually can simply be added because we know oxygen can form double bonds, so can carbon. Hydrogen will not form double bonds because it can only accommodate 2 electrons in its valence shell anyway. Notice now that the carbon has satisfied its valence, it has 8 electrons around it, the oxygen is deficient here. It doesn't have enough electrons yet. Furthermore we have only 2, 4, 6, 8 electrons in this structure, we're supposed to have 12. The remaining 4 electrons become lone pairs on the oxygen atom. Now we have a molecular structure in which the main group elements carbon and oxygen have octets, the hydrogens have their 2 pair of electrons, and the correct number of electrons, 2, 4, 6, 8, 10, 12 is equal to the number of electrons that we started off with. Now that we've got this system down, let's do another couple of molecules to make sure that we know what we're doing here. Next molecule we're going to consider is actually a molecule called Methonol. It has molecular formula CH4O. The number of available electrons is 4 for the carbon, 6 for the oxygen, and 4 for each of the hydrogens. The number needed is 8 for the carbon, 8 for the oxygen and 2 for each of the hydrogens, so that's 24. The number of shared electrons, clearly is 10, and the number of bonds that we expect to form is equal to 5. Let's see if we can figure out how to do that. Well I might begin by saying look I have four hydrogen's let's put them all around the carbon, but then there's no place for the oxygen. Instead we're only going to have three of the hydrogen's around the carbon. The carbon and the oxygen clearly have to be bonded to each other. That gives us four of the bonds, but we're still missing one of the hydrogen's which clearly must be on the oxygen. Now we have 1, 2, 3, 4 oxygens, all the atoms are at least present now. We have 1, 2, 3, 4, 5 bonds, which is the correct number of bonds. That's 2, 4, 6, 8, 10 electrons. There's still 4 electrons missing here. Where are those other 4 electrons? They are in pairs which complete the octet on the oxygen, leaving us with 2 electrons here on 2 pair a pair of electrons, 2 pairs of electrons. On the oxygen's, correct number of valence electrons, correct octet satisfied for each of the atoms present. Let's do one more example, before we end this lecture. We'll consider the very common molecule carbon dioxide. In this particular case, the number of available electrons is 4 for the carbon, plus 2 times 6 for the oxygens, that's 16. The number which are needed is equal to 8 for the carbon plus eight for each of the oxygen, that's 24. The number shared therefore is equal to 24 minus 16 is 8. The number of bonds therefore are equal to 4 . Let's draw our structure here. It turns out in carbon dioxide, the carbon is between the two oxygens. Will form the single bonds initially, that's only 4 of the available electrons and it's only 2 of the needed bonds. We can clearly make double bonds to each of the oxygens, such that we now lined up with all 4 of the bonds there. This is 2, 4, 6, 8 of the available electrons, but there are 16 available electrons. Carbon has satisfied its octet. There are 2, 4, 6, 8 electrons around the carbon, but the oxygens do not yet have their octets, but it all works out because the oxygens each need 4 more electrons. If we count up we have 2, 4, 6, 8, 10, 12, 14, 16 electrons which have been used. We have 2, 4, 6, 8 electrons around each oxygen and 2, 4, 6, 8 electrons around each carbon, so that the carbon dioxide molecule actually contains two double bonds. Also we were going to do one more. Let's do one more after this. Let's consider the hydrogen cyanide molecule, so we can do one that has nitrogen in it. The number of available electrons is equal to 4 for the carbon, 5 for the nitrogen, 1 for the hydrogens, so there are 10 available electrons, the number needed is 8 for the carbon 8 for the nitrogen, 2 for the hydrogen. That's 18. The number shared is therefore 18 minus 10 is equal to 7. The number of bonds is therefore equal to 4. Turns out in hydrogen cyanide, the carbon is between the nitrogen and the hydrogen. That's 4 of the electrons and 2 of the necessary bonds. I can't form any additional bonds between carbon and hydrogen, so I will have to form a triple bond between carbon and nitrogen. That's 2, 4, 6, 8 of the available electrons. There are two leftover electrons. But that's appropriate because the nitrogen at this point is still missing two electrons. So we'll add the remaining pair of electrons to the nitrogen atom. Now, carbon has 2, 4, 6, 8 electrons around it, nitrogen has 2, 4, 6, 8, electrons around it and there are 2, 4, 6, 8, 10 electrons in the structure consistent with the number we were supposed to start off with. This in fact the way in which we can routinely draw structures for molecules, and as we will discover over the course of the rest of the semester. Drawing these kinds of structures, gives us an enormous amount of energy. Not just about the stability of the molecule but about the strengths of the bonds. For example we can easily detect that the bond in this hydrogen cyanide molecule is quite strong. We'll learn about the geometries of molecules. We'll learn about molecular interactions. These structures in fact form much of the basis for all kinds of chemistry out there and we'll need to be able to draw these structures routine. In fact it is the heart of organic chemistry to be able to draw these structures. So we'll pick up more about these molecular structures in the next concept development study.
