Oxidation Numbers ```Name: Lori Status: educator Age: 40s Location: N/A Country: N/A Date: 2000 ``` Question: Is there a table that I can use to teach oxidation numbers, and what whould be the best method of memorizing them. Replies: Dear Lori, here is no complete table of a size convenient for memorization that I know of, although this information may be available online in some of the Periodic Tables on the web (for example the one at Berkeley). Many elements can exist in more than one oxidation state. Depending on the compound, N can be found in oxidation states ranging from -3 (in NH3) to +5 (in NO_3^-); S can be found from -2 to +6, and Cl can be found from -1 to +7! I teach my freshman chemistry students the following set of rules for determining oxidation states: "Oxidation numbers are counting rules used by chemists for the purpose of electron bookkeeping. They are used as if they are atomic charges, but they are usually not atomic charges at all. (1) atoms in pure elements are assigned an oxidation number of 0. Thus, Cl_2(g) and Fe(s) both have each atom's oxidation number = 0. (2) Monatomic ions are assigned an oxidation number equal to the charge on the ion. So the oxidation # of Fe^(2+) is +2 and the oxidation # of Cl^(-) is -1. (3) In a compound or polyatomic ion, F is assigned -1, H is almost always assigned +1, and O is nearly always assigned -2. These rules are given in the order of their priority. So in HF, F is -1 and H is +1. In H2O, H is +1 and O is -2. However, in H2O2 (hydrogen peroxide) H is +1 and O is -1. There are also ionic compounds where H is in the form of hydride (H-1). So one has to be careful, and remember that there may be exceptions in applying rule (3), and apply them in the given order of priority (F=-1, H=+1, O=-2). (4) The sum over oxidation numbers in a polyatomic molecule must always equal the total charge of the molecule (which is 0 if the molecule is neutral, + if it's a cation, and - if it's an anion). Thus, in VO_2^+, we have 1 = x + 2(-2), and x = +5 = vanadium's oxidation number. But in VO^{2+}, we have +2 = x -2, so x = +4. Vanadium can be prepared in many other oxidation states as well." I hope that this is helpful. This lets us figure out the oxidation number of Xe in xenon tetrafluoride, XeF4; 0 = x +4(-1), therefore Xe has an oxidation number of +4. Note that carbon atoms in organic compounds often have fractional oxidation numbers. Check out propane; C_3H_8; 0 = 3x +8 ; therefore x = 3/8 ! Nowhere do the rules say that oxidation numbers must be integers. best regards, prof. topper Click here to return to the Chemistry Archives

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