General Rules for assigning Oxidation Number to an atom

General Rules for assigning Oxidation Number to an atom

The following rules are employed for determining oxidation number of the atoms.

1.    The oxidation number of the element in the free (or) elementary state is always Zero.

Oxidation number of Helium in  He = 0

Oxidation number of chlorine in Cl2 = 0

2.    The oxidation number of the element in monoatomic ion is equal to the charge on the ion.

3.   The oxidaton number of fluorine is always - 1 in all its compounds.

4.    Hydrogen is assigned oxidation number +1 in all its compounds except in metal hydrides. In metal hydrides like NaH, MgH2, CaH2, LiH, etc., the oxidation number of hydrogen is -1.

5.    Oxygen is assigned oxidation number -2 in most of its compounds, however in peroxides like H₂O₂, BaO₂, Na₂O₂, etc its oxidation number is -

6.     Similarly the exception also occurs in compounds of Fluorine and oxygen like OF₂ and O₂F₂ in which the oxidation number of oxygen is +2 and +1 respectively.

7.    The oxidation numbers of all the atoms in neutral molecule is Zero. In case of polyatomic ion the sum of oxidation numbers of all its atoms is equal to the charge on the ion.

8.    In binary compounds of metal and non-metal the metal atom has positive oxidation number while the non-metal atom has negative oxidation number. Example. Oxidation number of K in KI is +1 but oxidation number of I is - I.

9.    In binary compounds of non-metals, the more electronegative atom has negative oxidation number, but less electronegative atom has positive oxidation number. Example : Oxidation number of Cl in ClF3 is positive (+3) while that in ICl is negative (-1).

Problem

Calculate the oxidation number of underlined elements in the following species.

CO₂, Cr₂O₇^2-, Pb₃O₄, PO₄^3-

Solution

1. C in CO_2. Let oxidation number of C be x. Oxidation number of each O atom = -2. Sum of oxidation number of all atoms = x+2 (-2) Þ x - 4.

As it is neutral molecule, the sum must be equal to zero.

\ x-4 = 0 (or) x = + 4

2. Cr in Cr₂O₇^2-. Let oxidation number of Cr = x. Oxidation number of each oxygen atom =-2. Sum of oxidation number of all atoms

2x + 7(-2)  =  2x - 14

Sum of oxidation number must be equal to the charge on the ion.

Thus, 2x - 14  =  -2

2x  =  +12

x  =  12/2

x  =  6