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I. Rule Number One
All free, uncombined elements have an oxidation number of zero.
This includes the seven diatomic elements (such as O2) and the other molecular elements (P4 and S8).
II. Rule Number Two
Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one)
III. Rule Number Three
Oxygen, in all its compounds except peroxides and superoxides, has an oxidation number of −2 (negative two).
IV. Rule Number Four
For single ions (in other words, not polyatomic), the charge on the ion is taken to be the oxidation number.
V. Rule Number Five
When no charge is indicated in a given formula, the total charge is taken to be zero.
With only a very few exceptions (which you will never see at an introductory level), oxidation states can be assigned to all atoms in a formula. There are more extensive sets of rules and, for the most part, they derive from the five rules above.
As an example, this rule is sometimes seen:
The oxidation state of fluorine is −1 in all of its compounds.
Here is an example. You might want to consider searching for redox rules to see how others have approached this topic.
Example #1: What is the oxidation number of Cl in HCl?
Since H = +1, the Cl must be −1 (minus one).
Example #2: What is the oxidation number of Na in Na2O?
Since O = −2, the two Na must each be +1. Notice that we treat each sodium as an individual unit, we do not address it as Na22¯.
Example #3: What is the oxidation number of Cl in ClO¯?
The O is −2, but since a −1 must be left over, then the Cl is +1.
Example #4: What is the oxidation number of Cl in ClO2¯?
Two O is −4 (from −2 x 2), but since a −1 must be left over, then the Cl is +3.
Example #5: What is the oxidation number of Cl in ClO3¯?
Three O is −6 (from −2 x 3), but since a −1 must be left over, then the Cl is +5.
Example #6: What is the oxidation number of Cl in ClO4¯?
Four O is −8 (from −2 x 4), but since a −1 must be left over, then the Cl is +7.
Example #7: What is the oxidation number of S in SO42¯
O = −2. There are four oxygens for −8 total. Since −2 must be left over, the S must = +6.
Example #8: What is the oxidation number of S in SO32¯
O = −2. There are three oxygens for −6 total. Since −2 must be left over, the S must = +4.
Example #9: What are the oxidation numbers in KCl?
K = +1 because K2O exists. The O is −2 by defintion, therefore each K must be +1 in order to keep the K2O formula at zero charge.
Cl = −1 because HCl exists. H is a +1 by definition, therefore Cl must be a −1. You can also say that Cl is a −1 because K must be a +1 and we need to have a zero charge for the formula.
Example #10: What is the oxidation number for each element in NaMnO4?
Na = +1 because Na2O exists. We know the O = −2, so each Na is a +1. (We could also use this: we know NaCl exists and that Cl is a −1. We know this because HCl exists. Therefore, the Na is a +1.)
O = −2 by definition
Mn = +7. There are 4 oxygens for a total of −8, K is +1, so Mn must be the rest.
Bonus Example: What is the oxidation number of N and of P in NH4H2PO4?
1) Let's split ammonium dihydrogen phosphate apart. It forms the following two polyatomic ions:
ammonium ---> NH4+
dihydrogen phosphate ---> H2PO4¯
2) Analysis of the ammonium ion:
We know that NH4+ takes on a +1 charge because we know that NH4Cl exists.
We know that the chloride ion is −1, therefore ammonium is a +1.
Remember that we know chloride is a −1 because HCl exists and H is defined as a +1 (except in hydrides).
In NH4+, the four hydrogens total to +4. (Ammonium is not a hydride.)
Therefore, the nitrogen is a −3, in order to leave a +1 for the overall charge on the ion.
2) Analysis of the dihydrogen phosphate ion:
Two H are present ---> +2
Four O are present ---> −8
In order for the entire ion to have a −1 charge, the P must be a +5.
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