How to calculate oxidation number and reduction number ? (In easy way)

Dear Student,

There is nothing known as reduction number. It is just oxidation number from which we can determine whether the substance is getting reduced or oxidised.
Oxidation number is a real or imaginary charge developed on an atom when it goes from its free state to combined state.

The oxidation number can be calculated according to the following rules
·  The oxidation state of a free element is zero
·  For a monoatomic ion, the oxidation state is equal to the net charge on the ion
·  In general, hydrogen has an oxidation state of +1 and oxygen has an oxidation state of –2 when they are present in most compounds (Exception: Hydrogen has an oxidation state of –1 in hydrides of active metalsand oxygen has an oxidation state of –1 in peroxides
·  The algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.
For example: the oxidation states of sulfur in
 H2S is calculated as (2×1) +x =0
  x = –2
The oxidation state of S8(elementary sulfur) is zero.

 

Regards,

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Oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom. Electropositive metal atoms, of group I, 2 and 3 lose a specific number of electrons and have always constant positive oxidation numbers.

In molecules, more electronegative atom gain electrons from a less electronegative atom and have negative oxidation states. The numerical value of the oxidation state is equal to the number of electrons lost or gained.

Oxidation number or oxidation state of an atom or ion in a molecule/ion is assigned by:

i) Summing up the constant oxidation state of other atoms/molecules/ions that are bonded to it and

ii) Equating, the total oxidation state of a molecule or ion to the total charge of the molecule or ion.

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