# 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.

## 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.