How to calculate oxidation number in Chemistry

How to calculate the oxidation number of elements in a compound is a major concern for some students all over the world but that shouldn’t be. The knowledge of oxidation state is applied in almost every aspect of chemistry; nomenclature of compounds,

Solving oxidation number and naming of compounds will be easier if we can learn how to deduce valency of elements.

Valency is defined as the number of electrons gained or lost in the formation of ions. Generally metals form positive valency while non-metals form negative valency.

The valency of metals is exactly the valence electrons, for example, Magnesium has atomic number 12 and the configuration is 2,8,2, therefore, the valency of the Magnesium atom is 2.

Valency of Some Metals Explained

Valency of Non-metals Explained

Oxidation number and naming of Acids

How to write the oxidation number of elements and name compounds

HNO₃ == calculate the oxidation number of Nitrogen and name the compound

1 + N + 3(-2) = 1 + N -6 = 0

N = 6-1 =5

The name of the acid is trioxonitate V acid

H₂SO₄==2(1) + S + 4(-2)

        S  == 2 + S -8 =0

        S =8 -2 = 6

The name of he acid s Tetraoxosulphate VI acid

H₂CO₃ == 2 + C + 3(-2)

       C  ==2 + C -6 = 0

       C = 6-2 = 4

Trioxocarbornate IV acid

Please remember that naming acids, there is no need to mention the hydrogen in the acid.

Oxidation number and naming of oxides

NO === N + (-2) = N-2 =0

N= 2

Nitrogen II oxide

NO₂ == N + 2(-2) =0

    N  ==N- 4=0

    N= 4

Nitroge IV oxide

N₂O = 2N -2 = 0

      2N = 2/2 =1

Dinitrogen I oxide

Oxidation number and naming of salts

Naming of Nitrates

KNO₃ = 1 + N + 3(-2) =0

             1 + N -6 =0

             N = 6-1 = 5

Potassium trioxonitrate V

Mg(NO₃)₂ = 2 + 2N + 6(-2) =0

                     2 + 2N -12 = 0

                     2N = 12-2 =10

                     N =10/2 = 5

Magnesium trioxonitrate V

NaNO₃ = 1 + N +3(-2) = 0

                1 + N -6 =0

                 N = 6-1 =5

Sodium trioxonitrate V

Al(NO₃)₃ = 3 +3N +9(-2)

               = 3 + 3N -18 =0

             3N = 18-3 =15

             3N =15

               N – 15/3 =5

Aluminium trioxonitrate V

Naming of Carbonates

CaCO₃ == 2+ C+ 3(-2) =0

               2 = C -6 =0

               C= 6-2 =4

Calcium trioxxocarboante IV

MgCO₃ == 2 + C + 3(-2)

             == 2 + C -6 =0

       C=6-2 =4

Magnesium trioxocarbonate IV

K₂CO₃ = 2(1) + C + 3(-2) =0

            == 2 + C -6 =0

             C == 6-2 =4

Potassium trioxocarbonate IV

Na₂CO₃ ==2(1) + C + 3(-2) =0

            = 2 + C -6 =0

             C == 6-2 =4

Al₂(CO₃)₃ = 2(3) +3C + 9(-2) =0

                == 6 + 3C -18 =0

                 3C = 18-6 =12

                 3C = 12 ; C = 12/3 =4

Aluminium trioxocarbonate IV

Naming of  Sulphates

MgSO₄ ==2 +S+ 4(-2)=0

             == 2 +S -8=0

              S = 8-2= 6

Magnesium tetraoxosulphate VI

CaSO₄ =2 + S + 4(-2) =0

           S = 8-2 =6

Calcium tetraozosulphate VI

K₂SO₄ =2(1) + S+ 4(-2) =0

        = 2 + S -8 =0

      S- 8-6 =6

Potassium tetraoxosulphate VI

Naming of Chromates      

 K₂Cr₂O₇ = 2(1) + 2Cr +7(-2) =0

    = 2 + 2Cr -14 =0

    2Cr =14-2 =12

    2Cr = 12

    Cr = 12/2 =6

Potassium heptaoxodichromate VI

Na₂Cr₂O₇ = 2(1) + 2Cr +7(-2) =0

    = 2 + 2Cr -14 =0

    2Cr =14-2 =12

    2Cr = 12

    Cr = 12/2 =6

Sodium heptaoxodichromate VI

Oxidation number and naming of complex ions

[V(H₂O)₆]³⁺ ==Hexaaquo Vandaium III ion

[Fe (H₂O)₆]²⁺ = Hexaaquo iron II ion

The oxidation state of the metal is outside the square brackets so there is no need to calculate it again.