Oxidation States

Oxidation States from Lewis Structures

The oxidation state of an atom gives us an indication of the electron density around the atom and it helps to keep track of the electron change in oxidation-reduction reactions. In the oxidation state formalism, we consider that each atom in a molecule or ion is bonded to the others through ionic bonds.

For each atom in the structure:
  1. Break all 2-electron bonds and give both electrons to the more electronegative of the bonding pair.

  2. Sum all electrons around the atom.

  3. Compare that number to the number of valence electrons of that atom.

    • total = number of valence electrons, oxidation state is zero

    • total > number of valence electrons, the atom has a negative oxidation state (-1 for every additional electron above the valence number)

    • total < number of valence electrons, the atom has a positive oxidation state (+1 for every electron below the valence number)

  4. Use Roman Numerals to indicate the oxidation state.

Oxidation-Reduction Half Reactions

Let's look at the transformation of NADP to NADPH and focus on the 3 carbon atoms and the nitrogen atom that seem to change their bonds (in red).

The oxidation state of the nitrogen atom doesn't change but 2 of the carbon atoms decrease in oxidation state. The carbon that bonds to the additional proton goes from -I to -II in oxidation state. One of the carbons bonded to nitrogen goes from I to 0 oxidation state. Overall, the addition of the 2 electrons causes the total of the oxidation states to decrease by 2 units.

In the conversion of water to 1/2 equivalent of molecular oxygen, we can see that the oxidation state of the oxygen atom changes from -II to 0. Two electrons are lost and the sum of the oxidation states of the atoms increases by 2 units.

Balanced Oxidation-Reduction Reaction

The net, balanced reaction has the increases in oxidation states of some atoms equal to the decrease in oxidation states of others.

Professor Patricia Shapley, University of Illinois, 2012