three F’s at −1 each. (Note that in each compound the sum of the oxidation
states is equal to zero.)- The oxidation state of oxygen is usually −2 in its compounds. Example:
H 2 O has two H’s at +1 each and one O at −2. (Exceptions occur when the
oxygen is bonded to fluorine and the oxidation state of fluorine takes
precedence, and in peroxide compounds where the oxidation state is
assigned the value of −1.) - The oxidation state of hydrogen in most compounds is +1. Examples: H 2 O,
HCl, and NH 3. (In hydrides, where hydrogen acts like an anion compounded
with a metal, there is an exception, however. In this case, hydrogen is
assigned the value of −1. Examples: LiH and KH.)
Some examples of determining the oxidation states of other elements in
chemical formulas follow.
Example 1
In Na 2 SO 4 , what is the oxidation state of sulfur?
The first thing to recognize is that this compound is an ionic substance. Ionic
substances have two parts—the cation and the anion. In this case, the cation is
monatomic and the anion is polyatomic. The cation is Na+ and the anion is SO 4 2−
(sulfate.) By Rule #2, the oxidation state of the sodium is +1 because the
oxidation states of monatomic ions are the same as their charges. By Rule #4, the
oxidation state of the oxygen in the sulfate is −2. Now you can look at the
complete formula and calculate the oxidation state of the sulfur.
Since the positive sum and the negative sum must equal 0,(+2) + x + (−8) = 0The sulfur must have a +6 oxidation state.Example 2
What is the oxidation state of carbon in the molecule CO 2?
By Rule #4, the oxidation state of oxygen is −2, and since there are two
oxygen atoms in this formula the total negative sum is −4. Since the positive and
negative sums must add to zero, the oxidation state of carbon is +4 in this