The oxidation bookkeeping method is a reliable way of determining the oxidation number of an atom. However, it usually requires electronegativity values and an electron dot diagram. Remembering electronegativity values by heart is difficult and drawing an electron dot diagram does take time. Because of this, chemists have devised a shortcut for assigning oxidation numbers. This shortcut does not apply to all compounds, but we can use it to assign oxidation numbers to many compounds we encounter in chemistry.
To use this shortcut, we must consider the following rules:
- The oxidation state of an atom in the most abundant form of an element is zero.
As a result, elements including the following have an oxidation number of zero:
Iodine: I2; Bromine: Br2; Aluminum: Al; Helium: He; Iron: Fe.
As you can see, elements including iodine, bromine, oxygen, chlorine, hydrogen are all diatomic. Recall that from the electron bookkeeping method, when two atoms of the same element share electrons, you must divide the shared electrons equally between them.
- The oxidation number of an oxygen atom in most compounds is -2. Except in peroxides, where oxygen has an oxidation number of -1. Examples of peroxides include hydrogen peroxides, H2O2(H—O—O—H).
- The oxidation number of elements in group one (alkali metals) of the periodic table is usually +1. Elements of this group include: hydrogen, sodium, potassium, and lithium.
- The oxidation number of elements in group two (alkaline metals) of the periodic table is usually +2. Elements of this group include: barium, magnesium, calcium, and sodium.
- The oxidation number of monatomic ions (one atom ion) is always equal to the charge of the ion. Examples of monoatomic ions include: Fe3+, Al3+, and O2-. From this list, you can see that the oxidation number of Fe3+is +3, and O2-is -2.
- In a chemical formula, the sum of the oxidation numbers of all the atoms it consists of must add up to the overall charge of the formula, even if the overall charge of the chemical formula is zero. For instance, for a polyatomic ion like CO32-, the sum of the oxidation numbers of C and O must add up to the charge of the ion, which in this case is negative 2, (-2). Similarly, for a neutral molecule like CO2, the sum of the oxidation numbers of C and H must add up to the overall charge of the ion, which in this case is zero.
- If a halogen (group 7 elements) is present in a chemical formula, then it’s most likely to have an oxidation number of -1.
As you can tell, these rules do not apply to all the elements on the periodic. However, you can combine them to determine the oxidation numbers of most elements in a chemical formula. For instance, what is the oxidation number of chromium (Cr) in the following chemical formula (dichromate ion):
Among the rules outlined above, we have no specific rule for Cr. But since the above formula is not a peroxide, we can conclude from rule 2 that the oxidation number of O is -2. And since we have 7 atoms of O in the formula, we must multiply 7 by -2. If we do, our answer will be -14. Also, since the overall charge of the formula is 2-, then it follows from rule 6 that the oxidation state of the chemical formula is -2. Now, from rule 7 again, we must ensure the sum of the oxidation numbers of atoms in the formula add up to the oxidation state of the formula. But since we don’t know the oxidation number of Cr, we will use the letter x to represent this number. But since we have two atoms of Cr in the formula, it follows that we must multiply x by 2. If we do, our answer will be 2x. Now, let’s translate all that into an algebraic expression. If we do, we will get something like this: