Chemistry 13.2 Oxidation Numbers - Welcome to Oxidation Numbers. In the last lesson, we saw that oxidation refers to the loss of electrons, and that reduction refers to the gain of electrons. Today, we're going to look at how to track this gain and loss of electrons-- what system we use for keeping track of the electrons that are transferred. And this is easy to see with ionic compounds because we know that if we have an atom of sodium and an atom of chlorine, the sodium has one valence electron, and the chlorine has seven. When I combine these two, we see the electron from the sodium is taken by the chlorine. It's a full transfer of electrons. So very clearly, the sodium is losing an electron, so it's oxidized. And the chlorine is gaining an electron, so it is reduced. Because the sodium and chlorine atoms gain a charge, it's very easy to see where the electrons went because they actually became ions. There's a full transfer of electrons. This looks a little bit different for covalent in terms of transfer of electrons because we know covalent refers to the sharing of electrons. So for example, hydrogen and chlorine. In this case, instead of the transfer of electrons, we have a shared pair. This electron is shared with this one between the two atoms. And that gives us our covalently-bound hydrogen chloride. However, the electrons in this bond are not evenly shared. Chlorine has a much higher electronegativity than hydrogen. So it essentially pulls the shared pair of electrons-- this bond-- it pulls it closer to the chlorine. Now earlier, we called this a polar molecule, and we showed it has a partial negative and a partial positive. Remember, that the partial negative came from the fact that the electron spent more time with the chlorine than with the hydrogen, or that they lived closer to the chlorine. So now we're going to look at that in a slightly different way. Instead of using these partial positive and partial negatives, we're going to say the chlorine is a minus 1, and the hydrogen is a plus 1, indicating that the chlorine has gained an electron and the hydrogen has lost an electron. Now these are not charges. The atoms don't actually get a official charge the same way they do in the ionic cases. But these do show us how the electrons move from one atom to another. And these are both called oxidation numbers. And oxidation numbers are simply a system used by chemists to keep track of what gains and what loses electrons as well as the number of electrons that are involved in that transfer. These are also sometimes called oxidation states. Now since these oxidation numbers are just a system, we actually have some rules we use to assign them to elements. So let's take a look at some of these rules for assigning oxidation numbers. So first, we're going to see that he oxidation number of a pure element is always zero. That means if I have carbon, or hydrogen, or oxygen, or phosphorous, or iron written like this, they have an oxidation number of 0. The second rule is that an ion of an element has the same oxidation number as its charge. So for the magnesium ion, it's oxidation state is going to be positive 2. And even for an ionic compound-- so an ionic compound like potassium iodide is made up of ions. We know it's made up of K-plus and I-minus. So the potassium ion here has a positive 1 oxidation state, and the iodine ion has a negative 1 oxidation state. And there's two side notes we can make about this rule as well that are really helpful to remember. And that's a group 1 ion-- so ions made a group 1 elements, like lithium, sodium, potassium-- they're always going to be plus 1 Whereas group 2 ions are always going to have a plus 2 charge or a plus 2 oxidation state. The next rule we're going to look at is a rule for the element fluorine. Fluorine is special because it's the most electronegative element. And it always has an oxidation state of negative 1. The next rule is about another element-- this time, hydrogen. And hydrogen is always going to have a +1 charge. So for example, when it's in the compound HCl or H2S, each of these hydrogens in these compounds has a +1 oxidation state. However, there is an exception for hydrogen. So when hydrogen is with a metal, it, instead, has a minus 1 oxidation state. So for example, sodium hydride-- sodium is a metal in this case, so hydrogen has a minus 1. And another example would be aluminum hydride. In this case, aluminum is the metal, and hydrogen has a minus 1 oxidation state. The next rule is about oxygen. And it's going to look similar to hydrogen in that we're going to have an exception. So oxygen is always minus 2. And there's actually going to be two exceptions for this one. So the first is except when it's a peroxide-- so except in peroxides, which are just a type of compound. And in terms of what we're going to come across, peroxides generally look like this. So H2O2 is hydrogen peroxide. Na2O2 is sodium peroxide. And in this case, it's minus 1. The second exception is when it's with fluorine. So there's no set number for this one, but we have to balance the fluorine. So these are the first five rules for assigning oxidation states. A pure element is always 0. The ion of an element is always the same as its charge. Fluorine is always negative 1. Hydrogen is always plus 1, with some exception. Oxygen is always minus 2 with some exception. And we'll use these five rules when dealing with the next two. Rule six says that the sum of all oxidation numbers in a compound is 0. Now the reason it's 0 is because compounds are neutral. They are neutral arrangements of elements. So let's look at the example of how to use this rule. Say, I take the compound KNO3 and the compound KNO2. So the first one's potassium nitrate. The second one's potassium nitrite. I can identify the potassium and oxygen just based off the rules we just used. So the potassium is going to be a plus 1. And the oxygen is going to be a minus 2. Same thing with the other compound-- potassium is a plus 1, and oxygen is a minus 2. But I don't know what N is. I have no rules for nitrogen. So I'm just going to put an x there. Now this rule tells me if I add up all the oxidation states, it has to equal 0. So let's see what that looks like for the first compound. The potassium is plus 1. I have to add that to the nitrogen, which is x. I have to add that to the oxygen, which is a minus 2. However, I have three oxygens. So that means I'm going to multiply this negative 2 by 3. And the rule says, if I add all of these up together, they have to equal 0. So now I can solve this for x to figure out what the nitrogen is. So I have 1 plus x minus 6 equals zero. So x has to equal positive 5. That means that this nitrogen that I did not know the oxidation number of, I was able to use this rule-- setting it equal to 0-- to find out that the nitrogen has a positive 5 oxidation state. Now it's going to be useful for us to go through the same process with the other compound. Again, I have a positive 1 from the potassium. I have to add that to x for the nitrogen. And then I have to add that to the negative 2 for the oxygen, except this time, I have two of them. So I multiply by 2. That gives me 1 plus x minus 4 equals 0. So in this case, x equals positive 3. So in this compound, potassium nitrite, nitrogen has an oxidation state of positive 3. And I was able to figure out both of these by using this rule, that the sum of all the oxidation numbers has to equal zero. And this also shows us something important, that some elements have multiple possible oxidation numbers. And so really, the only way for us to find out the oxidation numbers for things like nitrogen or other elements that have multiple possible oxidation numbers is by looking at them in a compound, using our other rules, and then setting the whole thing equal to 0. Now the next rule is similar to that one. This one says the sum of all oxidation numbers in a polyatomic ion is its charge. And this is going to be best to show with an example. So let's look at the chromate polyatomic ion, CrO4 two minus. And in this case, I don't know what the chromium is. I have no rule about chromium specifically. So that's an x. But I do know that oxygen is a minus 2 because it's not one of the exceptions. So I'm going to set this up-- x plus negative 2 times 4 oxygens, so there's four of them. And I set it equal to something else. I do not set it equal to 0 because a polyatomic ion, the rule says we're not setting it to 0. And that's because it's not neutral. It does have a charge. And the charge for this is the 2 minus. So that means I set this expression equal to negative 2 because the rule says that the sum-- so I'm adding all these up-- has to be equal to the charge. And then from this point, I just solve it as normal-- x minus 8 equals negative 2. So x equals positive 6. And that's going to be the oxidation number for chromium in this particular polyatomic. That wraps up our lesson on oxidation numbers and how to assign them. Write down any questions you have in your notes, and bring them with you to class.