Question

no one is answering me pls tell how to calculate oxidation no

Answer 1

The oxidation number of a free element is always 0.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of #"H"# is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of #"O"# in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Answer 2

Oxidation state or oxidation no is assigned according to a set of rules.

1. The oxidation no of  free element is zero
2. For mono atomic ion the oxidation no is equal to the charge on the ion.
3. The algebric sum of all the oxidation numbers in a free molecule is zero and in molecular ion it is equal to the net charge of the ion.
4.Hydrogen in compounds always have +1 except in metal hydrides it has -1.
5. Oxygen in compounds always have -2 except peroxides superoxides and fluorine compounds.
6. Fluorine has -1 in all its compounds
7.Alkali metals have +1 and alkaline earth metals have +2 in their compounds.

Example :

To find oxidation number of S in H2SO4

It is a free molecular compound.
Algebric sum of all oxidation numbers = 0
Hydrogen = +1
Oxygen = -2
Let oxidation no of S = x

2(+1) + x + 4(-2) = 0
2 + x - 8 = 0
or x = +6

Oxidation no of S in H2SO4 = +6

Example: Find oxidation number of S in H2S

2(+1) + x = 0
x = -2
Oxidation number of S in H2s is = -2

Example: Find oxidation number of Cr in Cr2O7^2-

Here it is a molecular ion having net charge of -2.
Therefore the sum of oxidation numbers = -2
Let the oxidation number of Cr = x

2x + 7 (-2) = -2
2x - 14 = -2
2x =  +12
x = +6

Practising more example will be helpful.