# Oxidation States

Oxidation states can be used to predict the compounds formed by a reaction.

Chemical bonding is related to the transfer and sharing of electrons. Generally, an atomic bond is described as a total transfer or equal sharing of electrons. Most bonds are made up of electrons that are unequally shared. To keep a record of electrons in compounds and during chemical reactions, scientists have devised a system called oxidation states. An oxidation state, also called an oxidation number, is a hypothetical charge assigned to an atom, ion, or polyatomic ion, indicating how many electrons have been lost (or gained). Negative oxidation states indicate greater access, and positive oxidation states indicate lesser access. Oxidation states are useful in understanding compounds as well as oxidation-reduction (redox) reactions. Oxidation states are related to electronegativity because electronegativity dictates the distribution of electrons in a molecule, which helps determine oxidation states. Strongly electronegative elements have a greater pull on electrons, depriving other atoms of them. This results in a negative oxidation state. While ions have actual electric charges, molecular atoms do not necessarily have charges. However, they can attract the electrons in a molecule in unbalanced ways. Oxidation states reflect this tendency, and electronegativity helps determine which atoms attract the electrons in a molecule.

The following are conventions regarding oxidation states:

• The oxidation state of an elemental molecule or atom is zero. A single neon atom, for example, has an oxidation state of zero. In hydrogen gas (H2), the oxidation states of both hydrogen atoms are zero.
• The oxidation state of an ion is equal to its charge. Thus, a sodium ion (Na+) has an oxidation state of +1, and a sulfur ion (S2–) has an oxidation state of –2.
• Nonmetals have greater electronegativity and therefore commonly have negative oxidation states. Metals have lower electronegativity and therefore commonly have positive oxidation states.
• Fluorine, the smallest halogen, is the most electronegative element on the periodic table. Fluorine always has an oxidation state of –1. Halogens other than fluorine also commonly have an oxidation state of –1, but exceptions do exist. The principal oxidation states of chlorine, bromine, and iodine are −1, +1, +3, +5, and +7.
• Oxygen is very electronegative. Oxygen commonly has the oxidation state –2. Exceptions to this are when oxygen is found in compounds with fluorine or in peroxide compounds. Oxygen has a very high electronegativity, so it almost always reduces, showing a negative oxidation state. Only fluorine shows higher electronegativity than oxygen. So in compounds containing both fluorine and oxygen, the fluorine has the ability to pull oxygen's electrons over, forming a positive oxygen ion. In compounds with fluorine, oxygen has weaker pull on electrons and therefore has a positive oxidation state. Only fluorine has the ability to make oxygen oxidize because only fluorine is more electronegative than oxygen.
• Alkali metals, which are group 1 elements except hydrogen, always have an oxidation state of +1.
• Alkaline earth metals, which are group 2 elements, always have an oxidation state of +2.
• Hydrogen usually has an oxidation state of +1 when it is bonded to nonmetals. When it is bonded to metals, it usually has the oxidation state of –1 because hydrogen is more electronegative than the metal it is bonded to.
• The sum of the oxidation states of all atoms in a neutral molecule is zero. The sum of the oxidation states of all atoms in an ion is equal to the charge of the ion. Thus, the sum of the oxidation states of all atoms in carbon dioxide (CO2) is zero, and the sum of the oxidation states of all atoms in a carbonate ion (CO32) is –2. This convention makes sure that the charge is conserved.

#### Ion Formation

Step-By-Step Example
Determining Oxidation States for Phosphorus Pentoxide (P2O5)
Phosphorus pentoxide (P2O5) is an oxygen compound, and the oxidation states of its atoms can be determined using conventional rules.
Step 1
Oxygen is not in a peroxide or coupled with fluorine here, so it has an oxidation state of –2. The sum of the oxidation states of all five oxygen atoms is –10.
Step 2
Phosphorus pentoxide (P2O5) is a neutral molecule, so its overall oxidation state is zero. The –10 that comes from the oxygen atoms must be balanced by the oxidation states of the phosphorus atoms.
Solution
With this information, it is possible to calculate the oxidation state of the phosphorus atoms. Use the variable x to represent the oxidation state of phosphorus. The following relation must hold true:
\begin{aligned}2x& = 10\\x& = 5\end{aligned}
Thus, in phosphorus pentoxide (P2O5), each oxygen atom has an oxidation state of –2, and each phosphorus atom has an oxidation state of +5. These numbers mean that in phosphorus pentoxide (P2O5), oxygen has more access to electrons than phosphorus does.
Step-By-Step Example
Determining Oxidation States for Sulfate Ion (SO42–)
Use the rules for oxidation states to determine the oxidation states of a sulfate ion (SO42–).
Step 1
Each oxygen atom will have the oxidation state of –2. The sum of the oxidation states of four oxygen atoms is –8.
Step 2
A sulfate ion (SO42−) has an overall charge of 2–. Its overall oxidation state will be –2. The –8 oxidation state that comes from the oxygen atoms must be reduced to –2.
Solution
There is one sulfur atom. It must have an oxidation state of +6 because $8+6=-2$.
\begin{aligned}x+4\left({-2} \right)& =-2\\x& =-2+8\\x &=+ 6\end{aligned}