The oxidation number, sometimes called the oxidation state, represents the hypothetical charge an atom would possess if all of its bonds with other atoms were purely ionic. This means the electrons in each bond were completely transferred to the more electronegative atom. This assigned number helps track the movement of electrons during chemical reactions. By tracking the change in oxidation numbers, one can easily identify which atoms are losing electrons (oxidation, indicated by an increase in number) and which are gaining electrons (reduction, indicated by a decrease in number).
Establishing the Fixed Oxidation Number Rules
The trick to finding an unknown oxidation number lies in first knowing the fixed values for certain common elements. Any element found in its elemental or uncombined form, such as O2, S8, or pure Fe metal, always has an oxidation number of zero.
When elements are combined in a compound, the most reliable numbers are assigned first. Fluorine (F) is the most electronegative element and always has an oxidation number of -1 in all of its compounds. Metals from Group 1, like lithium (Li) and sodium (Na), are consistently assigned +1. Group 2 metals, such as magnesium (Mg) and calcium (Ca), are fixed at +2.
Hydrogen (H) is generally assigned an oxidation number of +1 when bonded to a nonmetal. The exception occurs when hydrogen is bonded to a metal in a metal hydride, such as NaH, where it takes on an oxidation number of -1. Oxygen (O) is usually assigned a value of -2 in compounds. However, in peroxides (like H2O2), the oxidation number of oxygen changes to -1.
Calculating Numbers in Neutral Compounds
Once the fixed numbers are established, finding an unknown oxidation number in a neutral compound becomes a straightforward algebraic process. The fundamental principle is that the sum of the oxidation numbers for every atom in a neutral compound must equal zero. This rule allows for the determination of the oxidation number for an element that does not have a fixed value, such as sulfur in sulfuric acid (H2SO4).
To find the oxidation number of sulfur (S), we first assign the known values for hydrogen (+1) and oxygen (-2). The calculation is set up by multiplying the oxidation number of each element by the number of atoms present in the formula, and then summing them to zero.
For H2SO4, the equation is 2(ON of H) + 1(ON of S) + 4(ON of O) = 0. Substituting the known values gives 2(+1) + x + 4(-2) = 0, where x represents the unknown oxidation number of sulfur. Solving this equation yields x – 6 = 0, resulting in an oxidation number of +6 for the sulfur atom.
Calculating Numbers in Polyatomic Ions
The method for calculating oxidation numbers in polyatomic ions is nearly identical to that for neutral compounds, but the final sum differs. Since a polyatomic ion carries a net electrical charge, the sum of all individual oxidation numbers within the ion must equal that net charge, rather than zero.
For example, to find the oxidation number of sulfur in the sulfate ion, SO4(2-), the calculation is set equal to the ion’s charge of -2. The known oxidation number for oxygen is -2, and there are four oxygen atoms present. The algebraic expression is written as 1(ON of S) + 4(ON of O) = -2. Letting x be the oxidation number of sulfur, the equation becomes x + 4(-2) = -2. Solving this yields x – 8 = -2, resulting in +6.