 Today on Flipping Science, we're looking at finding oxidation states and oxidation numbers. This is a handy thing to do because it tells you how many electrons are being lost, gained, or shared with other atoms in polyatomic compounds. It also gives you a hint about what's going to happen in terms of redox reaction. Some periodic tables, they have the oxidation states on there. So here we have, for example, nitrogen to plus minus 3, 4, and 5, fluorine and oxygen, fluorine minus 1, oxygen minus 2, and we'll see where that comes from in a second. So there's a series of rules that we follow to figure out the oxidation number of a substance. We'll go through these in order. So the oxidation number of neutral atoms or molecules of an element is equal to zero. That means something like, say, nitrogen gas, the total oxidation number would be zero. Oxygen gas, the oxidation number would be zero. All right, the oxidation number of a monatomic ion equals the charge of that ion. So if you had Fe plus 3, the oxidation number of that one would be plus 3. In neutral molecules, the oxidation numbers add up to zero. So if we add H2O, the oxidation numbers there have to add up to zero because there's no charge up here. If we were looking at polyatomic ions, it has to equal the charge on that ion. So the oxidation number has to equal the charge on that ion. So if we had, say, sulfate, SO4, 2 minus, the oxidation number there would be equal to minus 2. Then we're going to very specific examples. So fluorine is always minus 1, oxygen is always minus 2, except in peroxides. Group 1 ions are always plus 1, group 2 ions are always plus 2, halogens are always minus 1, and hydrogen is always plus 1 unless it's with electronegative elements down there. So let's look at some examples of the questions that you're likely to get. So here's a question, find the oxidation number of sulfur in sulfuric acid. The way to do this is basically year 9 level mass. So we're going to call our unknown X up here, and we've got numbers of the other bits, so two hydrogens, four oxygens, and we know their oxidation number from the rules. So what we're going to say is two times the oxidation number of hydrogen. Now if we go back to our rules, we can see hydrogen is always plus 1, unless it's with those, it's not really with those, so in this case it's going to be plus 1. So two times plus 1, plus we've got one sulfur is equal to X plus we've got four oxygens, so four times the oxidation number of oxygen. If we look at our rules, the oxidation number of oxygen is always minus 2, unless it's in a peroxide, and it's not in this case. So four times minus 2. That has to equal the oxidation of the charge on the complete thing, and we see there's no charge up here. So in this case that has to equal zero, and then we just solve for X. So two plus X minus eight equals zero. So two minus eight is minus six, so X minus six equals zero, so then X equals plus six. Make sure you always put the sign. That's very important. All right. So in that case the oxidation number of sulfur in sulfuric acid is plus six, and the way we represent that is we draw that on top. So plus six up there. Let's look at another example. Find the oxidation number of P in this molecule here, and notice there's a charge now. So we're going to do the same thing. So two times plus one, because that's the oxidation number of hydrogen here, plus X, which is our phosphorus that we're looking for, plus four times minus two for the oxygens that we had there. This time rather than equaling zero, it's going to equal minus one. So we've got two plus X minus eight equals minus one. So two minus eight is minus six again, so X minus six equals minus one. So X equals plus five. So the oxidation number of phosphorus there is plus five, so we write that on top.