Laster

Video transcript

- [Voiceover] In this video, we're going to find the oxidation state of carbon in several different molecules. In earlier video, we've already seen the definition for oxidation state, and also how to calculate it. So let's start with methane, and let's find the oxidation state of carbon and methane. One approach is more of a general chemistry approach where we know that hydrogen usually has an oxidation state of plus one, and we have four hydrogens for a total of plus four. The sum has to be equal to zero, so we know that carbon's oxidation state must be minus four immediately, since we only have one carbon here. So let's go ahead and verify that with our dot structure. So remember, when we're calculating the oxidation state using dot structures, we're thinking about bonding electrons, and we know that each bond consists of two electrons, so we need to put in the bonding electrons for all of our bonds. Next, we think about the oxidation state for carbon, and we start with a number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have, and we know carbon is supposed to have four valence electrons, so from that number, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons carbon has in our drawing. But now we need to think about these covalent bonds as being ionic, and so the more electronegative atom is going to take all of the electrons in the bond, so we need to think about electronegativity differences, and we're comparing carbon to hydrogen. So which is more electronegative? We know that carbon is more electronegative than hydrogen, so the two electrons in this bond here, carbon is going to take both of them, so it's winner takes all. Carbon's going to hog those electrons in this bond. All right, same for this next carbon-hydrogen bond. Carbon is more electronegative, so it takes those electrons, and all the way around. So we can see that carbon is now surrounded by eight electrons right, let's count them up here, one, two, three, four, five, six, seven, eight, so four minus eight is equal to minus four, so we already knew that minus four was going to be the oxidation state for carbon. Let's move on to another molecule here, so C2H4, this is ethene, or ethylene. What's the oxidation state of carbon in this molecule? Well hydrogen should be plus one, and we have four of them for a total of plus four. So the total for carbon should be minus four, because that total has to sum to zero, but this time we have two carbons, so minus four divided by two gives us minus two, each carbon should have an oxidation state of minus two. And let's verify that, let's put in our bonding electrons, and let's calculate the oxidation state of carbon by using our little formula here. So you put in our bonding electrons, and let's just pick one of the carbons to start off with, let's say we're talking about the carbon on the right. So we think about electronegativity differences, and we know carbon is more electronegative than hydrogen, so carbon's going to steal those electrons, and same thing over here, carbon's more electronegative, so carbon steals those electrons too. When we get to this double bond here between the two carbons, we have these four electrons, and now we're trying to compare the electronegativity of carbon, to carbon, and obviously, that's the same electronegativity. So with four electrons, we're going to divide those electrons equally, because both of those carbons have the same value for the electronegativity and so we take those four electrons, we divide them in half, so we give two electrons to one carbon and two electrons to the other carbon. So carbon normally has four valence electrons, right, so we're using our formula here for oxidation state. Oxidation state is equal to the number of valence electrons that carbon is supposed to have, minus the number of valence electrons around carbon in our drawings, so let's count them up after we've accounted for electronegativity. That's one, two, three, four, five, and six, so four minus six gives us an oxidation state for carbon of minus two, which is in agreement with what we calculated over here. Keep in mind that each carbon was supposed to have an oxidation state of minus two, so if we look over here at the carbon on the left, and we just assign those electrons really quickly, we can see that that would be the same caluclation, four minus six gives us minus two, so each carbon has an oxidation state of minus two. We move on to another molecule, so CH2O, this is formaldehyde, and we know that oxygen usually has an oxidation state of minus two, and we have one oxygen for a total of minus two down here. Hydrogen usually has an oxidation state of plus one, and we have two of them for plus two. This total has to sum to zero, so carbon should have an oxidation state of zero in this molecule. And let's go ahead and look at the dot structure. We put in all of our bonding electrons, so we can find the oxidation state easier, and we think about electronegativity differences. We know carbon's more electronegative than hydrogen so carbon takes those electrons. Same for this bond, carbon takes those electrons. But now, now we're going to compare carbon to oxygen, and we know oxygen is more electronegative than carbon, so oxygen's going to take all of those electrons, so all four of those electrons in that double bond. So it's winner takes all when you're thinking about oxidation states. So because oxygen is more electronegative, it takes all of those electrons, and now we can see that carbon is surrounded by four electrons, so one, two, three, four, carbon is supposed to have four valence electrons around it, and here we see it is surrounded by four and once we take into account electronegativity, so four minus four gives us an oxidation state of zero, so in the formaldehyde molecule, carbon has an oxidation state of zero, which is what we predicted over here. Let's think about oxygen. We said that oxygen would have an oxidation state of minus two, so let's look at oxygen, oxygen is right here. Let's think about how many valence electrons oxygen should have. Well, because of it's position on the periodic table, oxygen should have six valence electrons. How many electrons are around it now, once we've accounted for electronegativity? Well, here's one, two, three, four, five, six, seven, and eight. So six minus eight gives us an oxidation state for oxygen of minus two. What about for hydrogen? So hydrogen should have one valence electron, but there's zero electrons around it here because carbon's more electronegative than hydrogen, so one minus zero gives us an oxidation state of plus one, which is what we predicted over here, and the same thing for this hydrogen. So this formula works for other atoms too. All right, let's do another example, and this time we're doing formic acid, so CH2O2, oxygen should be minus two, and we have two of them, so we have minus four for totaling our oxygens. Hydrogen should be plus one, and we have two of them for a total of plus two, and now, this total has to (mumbles) equal zero. So we know that carbon's oxidation state should be plus two in the formic acid molecule here. So let's go over and put in our bonding electrons, so we put in all of our bonding electrons here, and we think about the oxidation state of carbon, we think about electronegativities. So carbon is more electronegative than hydrogen, so carbon gets those two electrons in that bond, but over here, oxygen is more electronegative than carbon, so oxygen takes those electrons, and the same thing above, oxygen is more electronegative than carbon, so oxygen takes those electrons too. So what's the oxidation state of carbon in formic acid? Well, carbon is supposed to have four valence electrons, and from that we subtract the number of electrons around carbon, once we've accounted for electronegativity. Once we pretend like everything is ionic, and that's only two electrons this time, so four minus two gives us an oxidation state for carbon of plus two, just like we predicted over here. So you can see every molecule we've done so far, has had a different oxidation state for carbon, carbon is unique, it has all these different oxidation states. Let's finally look at carbon dioxide. So oxygen should have an oxidation state of minus two, and we have two of them for a total of minus four, so carbon must have an oxidation state of plus four, so we look at our dot structure over here, and we put in our bonding electrons, and we think about electronegativity. Oxygen is more electronegative than carbon, so oxygen is going to steal -- each oxygen is going to steal the electrons in those bonds. Let me re-draw it here. I'll show oxygen with all these electrons around it so the four electrons in that double bond here, are all going to be on this oxygen on the left. Same thing for the oxygen on the right. It has these lone pairs of electrons, and it's more electronegative than carbon, so it steals all of those electrons, which leaves carbon with zero electrons around it, so to find carbon's oxidation state, we know carbon should have four valence electrons, and here, we have zero electrons around carbon. So four minus zero gives us an oxidation state of plus four, just as we predicted, and if you wanted to do each oxygen, oxygen should have six valence electrons, and here we have one, two, three, four, five, six, seven, eight, so six minus eight gives us minus two for the oxidation state of oxygen. All right, let's go back up to the beginning here. Let's look at all these different oxidation states that we covered for carbon. So we started off with an oxidation state for carbon of minus four, we went to minus two, and then we went to zero, and then we went to plus two, and then finally, we ended up with plus four. So carbon can have a range of oxidation states, from minus four to plus four, when you're talking about carbon with four bonds. And you can also have in between, right, it's possible, of course, for carbon to have an oxidation state of plus three, I just didn't do an example like that in this video.