Calculating Oxidation Numbers

In this exercise, we're going to determine the oxidation number of sulfur in different compounds. So, we are asked to determine the oxidation number of sulfur in two molecular species. The first one is "A" and the second one is "B" in the elemental form, which is "C." We are also asked to determine the oxidation number of sulfur in two substances containing ions, "D" and "E." In each species, the sum of the oxidation numbers of all the atoms must equal the charge on the species. So, in this case, the total oxidation number is zero as it is a neutral compound. The same applies to these compounds, which are also neutral. In compound "B," the total sum of oxidation numbers is -2. Okay, so we will use the rule outlined previously to assign the oxidation numbers. So, first of all, when hydrogen is bonded to iron metal, its oxidation number is +1. As we can see in exercise and the sample example A, we have H2S and hydrogen is +1. Since H2S is overall neutral, the sum of the oxidation numbers must equal zero. So, if this is +2, that means S is -2. Thus, we have the equation 2x + 1 (because we have two hydrogens and each is +1) equals -2. To solve for x, we need to rearrange and subtract 2 from each side. Thus, x equals -2. So, in the first one... Sulfur has an oxidation number of minus two in its elemental form. Therefore, the oxidation number for sulfur in this form is equal to zero. This is true for every atom, resulting in a total of zero for the entire compound. Now let's consider SCl-. This is a binary compound and chlorine is from family 7, which means it has an oxidation number of minus 1. Since there are two chlorine atoms in this compound, the equation becomes X - 2 = 0, where X represents the oxidation number of sulfur. In a way that we have the X by itself, we need to add 2 on each side. That will make X equal to +2. Okay, so in that way, here the oxidation number for sodium is +2, here it is -2, and here it is 0. Let's go to see the example of sodium sulfide. In this compound, sodium is from family 1 and it has an oxidation number of +1. We have two sodium atoms, so the total oxidation number for sodium is +2. On the other hand, oxygen is from family 6 with an oxidation number of -2, and we have three oxygen atoms, so the total oxidation number for oxygen is -6. Therefore, the equation becomes: 2 - 6 + X = 0. To rearrange this equation, we will add +4 on each side, so that means that the sulfur will have an oxidation number of +4. Okay, we have 2 is because of the sodium, each one is +1, so we have two of them. Two + (2x + 1) will be +2. We have three oxygen, and each oxygen has an oxidation number of -2. We have 3, so it can be -6 + x. That is the oxidation number of sulfur, which is going to be equal to 0 because the net oxidation number of the whole molecule is going to be 0. Okay, and we rearrange this to find the value of x, and that's going to be 4. +4 in the example we have SO4, -2 in aspiration oxyanion S, -2. We have four of them, so this is going to be -8. But now it's going to be x - 8 = -2. Remember that. Here, the iron has a -2 charge, so that means that this must be equal to -2. So, to do that, this X must be a +6. +6 - 8 is going to be -2, and that's the oxidation number of sulfur. And this is in the sulfate ion.