Use of Oxidation Numbers |
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An ion is a charged atom or group of atoms. A monatomic ion consists of one atoms, while a polyatomic ion consists of two or more atoms. The charge of a ion is called its oxidation number. Oxidation numbers are very important in much of chemistry because many times atoms do lose or gain electrons. When this happens, they become ions. However, oxidation numbers are especially important when writing chemical formulas for ionic compounds. The following is a step-by-step process of how one would write a chemical formula.
Step 1: Look up the oxidation numbers for both sodium (Na) and chlorine (Cl) on a table. You will see that the charge for Na is 1+, while the charge for Cl is 1-. Step 2: Combine the ions so that the sum of the charges equals zero. In this case the charges of the ions, when added together, naturally equal zero. Therefore, no subscripts need be added. 1 + (1-) = 0. Step 3: Having checked the oxidation numbers and made sure they add up to zero, you can now just put the Na and the Cl together to form NaCl. Writing chemical formulas is actually not a very complicated process. It is pretty straightforward. However, the following example is a bit trickier because the use of a subscript comes into play.
Step 1: Look up the oxidation numbers for both the calcium (Ca) ion and the chloride (Cl) ion. Note that chloride is just a name for the monatomic ion formed from chlorine. You will see that the charge of Ca is 2+, while the charge of Cl is 1-. Step 2: Combine the ions so that the sum of the charges equals zero. Here, however, a subscript must be used. 2 + (1-) does not equal 0. Therefore, the subscript 2 must be added to the Cl ion, because 2 + 2(1-) = 0. Step 3: Having checked the oxidation numbers and made sure they add up to zero, you can now just combine the Ca and the Cl2 to form CaCl2.
Step 1: Look up the oxidation numbers for both of these polyatomic ions. Ammonium, you will see, has a +1 charge (NH4+). Dichromate, you will see, has a 2- charge (Cr2O7 2-). Step 2: Combine the ions so that the sum of the charges equals zero. Here, however, a subscript must be used. Also, parentheses must be used because you are using polyatomic ions. You will need to add a subscript 2 to NH4 so that the sum of the charges equals zero. Step 3: Place parentheses around the NH4 so that the whole NH4 has the subscript 2. Add on the Cr2O7. Your final result will be (NH4)2Cr2O7. Here's a list of a few other pages on oxidation numbers:
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