How to Calculate Oxidation state

Oxidation and reduction are two processes encountered during chemical reactions. Generally, oxidation involves the loss of electrons, while reduction means the gain of electrons. During this loss and gain of electrons the electronic distribution on atoms in a molecule changes. Oxidation state or oxidation number therefore is the degree of loss of electrons by an atom.

We can calculate the oxidation number of atoms in a molecule by analyzing its electronic state. It is explained by a few examples here.

Carbon dioxide (CO₂)

Carbon dioxide has a carbon atom doubly bonded to two oxygen atoms. The oxidation state of carbon can vary from compound to compound. Oxygen belongs to group 16 having six electrons in the valence shell and a valency of -2. Two oxygen atoms would have a total charge of -4. As the molecule is neutral so the overall charge must be zero. So the oxidation number of carbon should be +4 to make the overall charge zero.

So the oxidation state of carbon in carbon dioxide is +4.

Let’s look at this from another perspective. Oxygen has six electrons in valence shell. In carbon dioxide each oxygen has a total of 8 electrons. Two electrons shared by the carbon are counted with oxygen because it is more electronegative than carbon and the electron density is inclined towards it. Therefore it has an excess of 2 electrons, and a -2 charge. Or 6-8 = -2

Carbon on the other hand does not have any electrons in carbon dioxide, because it is attached to two highly electronegative atoms (O), so the shared electrons are counted with oxygen not carbon. Normally carbon bears 4 electrons in valence shell. Therefore, C in CO2 is deficient of 4 electrons and thus bears +4 charge. Or 4 – 0 = +4.

 Acetic Acid

Acetic acid has two carbon atoms. Let’s calculate the oxidation number of both. The methyl carbon is attached to three hydrogens which are less electronegative than carbon. So from the three bonds with hydrogens carbon has a total of six electrons. The fourth bond with the carbonyl carbon is shared equally to give one electron to each carbon. Therefore the methyl carbon has a total of 7 electron. As it normally bears 4 electrons, it is in excess of 3 electrons here, therefore a -3 charge. Or 4 – 7 = -3.

The carbonyl carbon is attached to a carbon and two oxygen atoms. It is less electronegative than oxygen so it doesn’t get any electrons from the bonds with oxygens, while its gets one from the C-C bond. Therefore it has a total of 1 electron in contrast to what it has (4) normally. It is deficient of 3 electrons therefore +3 charge. Or 4 – 1 = +3.

Potassium permanganate (KMnO₄)

The oxidation state of Mn in KMnO₄ is +7. Let’s see how.

Potassium is in group 1 with an oxidation state of +1. Oxygen has an oxidation state of -2. Four oxygens combined has a charge of -8. As the molecule is neutral, the oxidation state of Mn should be such that the total charge becomes 0.

Therefore,    +1 + X + (-8) = 0

                  X – 7 = 0

So,                X = +7

 Sulphate anion (SO4^-2)

To calculate the oxidation of number of S. Oxygen has a combined charge of -8. The ion has a total charge of -2.

Therefore, X + (-8) = -2

Or             X = 8 -2 = +6

The oxidation state of S thus is +6.

PC: www.answers.com