# Oxidation number of an element...

The oxidation number (o.n) is an integer ( can be positive or negative )represented in roman numbers (for e.g :IV ) and the role of this number is to help us in balancing reactions and identifying it as an oxidation/reduction reaction.

In conventi0n , we have rules for identifying the oxidation number for elements which are :

**1) The oxidation number of an element is equal to zero (O.N = 0 ) .**

Examples : – Fe O.n=0

- N_{2} O.n=0

- O_{2} O.n= 0

**2) The oxidation number (o.n) of a mono-atomic ion is equal to the charge of the ion .**

Examples : – H^{+} , O.n=+**I**^{
}

- N^{3−} , O.n = -**III**

**3) The oxidation number of grouped oxygen is O.n= -II except in peroxides (H_{2}O_{2} and N2O_{2 )} the oxidation number of oxygen becomes O.n= -**

**I**

Example : – in NaO , O.n(O)= -**I****I**

- in **H _{2}O_{2}** , O.n(O) = -

**I****4) The oxidation number of grouped Hydrogen is O.n= + I except for hydrides ( NaH and LiH ) The oxidation number of Hydrogen is O.n=-**

**I**Example :

- in HCl , O.n(H) = +**I**

- in LiH , O.n(H) = -**I**

**5) The oxidation number of grouped elements from group 1 in the periodic table is O.n =+I and that of group two is O.n = +II **

- in NaCl , O.n(Na) = +**I**

- in CaO , O.n(Ca) = +**I****I**

**6) In a molecule the sum of oxidation number of atoms constituting the molecule is equal to zero **

Example : NO2 , O.n(N) + 2[O.n(O)] = 0

O.n(N) -2 = 0

O.n(N) = +**I****I**

**7) In a polyatomic ion the sum of oxidation number of the ions is equal to the whole charge .**

Example : NH_{4}^{+}

O.n(N) + 4[O.n(H)] = +**I**

O.n(N)+4(+1) = +**I**

O.n(N)=+**I**-4

O.n(N)= -**III**

**8) The oxidation number of : – Grouped Al = +III**

** – Grouped F = -I**

This was a brief summary about oxidation number and the rules , if you have any questions do not hesitate in commenting or contacting me !

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