Oxidation Number and Concept of Oxidation and Reduction:

The definition of oxidation and reduction based on loss or gain of electrons is limited to the scope. The definition holds goods for ionic compounds.
 Examples, the formation of water from hydrogen and oxygen, can not be covered by the electronic concept since water is not an ionic compound.

2H_{2} + O_{2} → 2H_{2}O

It may recall classically we could still say that hydrogen is oxidized to H_{2}O. In the same seance burning of magnesium in oxygen is considered oxidation.
Similarly, hydrogen and chlorine react to form a covalent molecule hydrogen chloride.

2H_{2} + Cl_{2} → 2HCl

The Above reaction hydrogen is oxidized or chlorine is reduced but the resulting compound is covalent one, thus the reaction cannot be covered by the electronic concept.
In order to cover such reactions also under oxidationreduction, the concept of oxidation number is developed and it is defined as,
Oxidation Number:

The oxidation number of an element in a compound is the formal charge (positive or negative) which would be assigned to the element if all the bonds in the compounds were ionic bonds.

All the compounds are treated as though they were ionic merely because of the case of counting oxidation numbers. The oxidation number, therefore, is arbitrary.

Electronegativity concept is utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound is arbitrarily assigned a positive oxidation number and more electronegative one a negative oxidation number.

Florine, being the most electronegative, has always a negative oxidation number. The alkali metals are highly electropositive, have low electronegativity, hence have been assigned positive oxidation numbers.
Method of finding out the Oxidation Number:
The following general rules are to be observed for the assignment of oxidation numbers.
 1. Atoms of diatomic molecules like H₂, Cl₂, O₂, etc or of metallic elements are assigned zero oxidation numbers since the same elements of similar electronegativity are involved in the bonding.

H ➖H

The oxidation number of the above molecules are zero because two hydrogen atom of the same electronegativity is involved for bonding.
 2. Except for metal hydrides, the oxidation number of hydrogen in hydrogen compound is +1. In alkali metal hydrides, LiH, NaH, CsH, etc, the oxidation number of hydrogen, for reasons of varying electronegativity, is 1.

NaH → Naᐩ + H⁻
(Here oxidation number of H is 1)
HCl → Hᐩ + Cl⁻
(Here oxidation number of H is +1)
 3. The oxidation number of metal is positive.

CuO → Cuᐩ² + O⁻²
(Here Oxidation number of Cu is +2)
 4. Oxygen has normally an oxidation number 2. In hydrogen peroxide, the oxidation number of oxygen is 1 since hydrogen has to be assigned +1.

In peroxides and superoxides, the oxidation number of oxygen is 1 and 1/2 respectively.

In fluorine monoxide(F₂O) oxygen has an oxidation number +1 because fluorine is more electronegative than oxygen.

For Examples, CuO → Cu+2 + O2
(Here Oxidation number of O is 2).

In H₂O, the oxidation number of Oxygen is (2), but in H₂O₂, the oxidation number of H is (+1) and oxidation number of Oxygen is (1).

In BaO₂, the oxidation number of Ba is (+2) and thus the oxidation number of Oxygen is (1).

In Na₂O, the oxidation number of Na is (+1) and the oxidation number of oxygen is (2).
 Problem 1:

Oxidation number of P in Ba(H_{2}PO_{2})_{2} is  (a)+3, (b)+2, (c) +1, (d) 1.
 Answer:

(c) +1
 The oxidation number of Ba is +2, the oxidation number of hydrogen is +1 and the oxidation number of oxygen is 2.

Let the oxidation number of P is x.
∴ (+ 2) + 2{2(+1) + x +2(2)} = 0
or, 2x  2 = 0
or, x = +1
 5. The oxidation number of an ion is equal to its charge.

NaCl → Na⁺ + Cl⁻

The charge and oxidation number of Na+ are +1 and the charge and oxidation number of Cl⁻ are 1.

MgBr₂ → Mg⁺² + 2Br⁻

Here the charge of Mg⁺² is +2 and the oxidation number also +2 and the charge of Br⁻ is 1, thus the oxidation number also 1.
 6. The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero.

In HCl oxidation number of hydrogen is +1 and the oxidation number of chlorine is 1.
And the sum of these = (+1) + (1) = 0.
Oxidation Number of an Element in a compound:
Oxidation Number of Mn in KMnO₄:

Let the oxidation number of Mn in KMnO₄ is x.
Thus according to the above rule,
(+1) + x + 4(2) = 0
or, x = +7
Thus, the oxidation number of Mn in KMnO₄ is +7.
Oxidation Number of Mn in MnO₄⁻²:

Let the oxidation number of Mn in MnO₄⁻² is x and the oxidation number of oxygen is 2(according to the above rule).

Thus the sum of the oxidation number of MnO₄⁻² = Charge of the MnO₄⁻².
∴ x +4 (2) = 2
or, x = +6
Thus, the oxidation number of Mn in MnO₄⁻² is +6.
Oxidation Number of Cr in Cr₂O₇⁻²:
Let the oxidation number of Cr in Cr₂O₇⁻² is x
∴ 2x + 7(2) = 2
or, x = +6
Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6.
∴ 2x + 7(2) = 2
or, x = +6
Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6.
Oxidation Number of S in H₂SO₄:
Let the oxidation number of S in H₂SO₄ is x. According to the rule oxidation number of hydrogen is +1 and oxygen is 2.
∴ 2(+1) + x + 4(2) = 0
or, x = +6
Thus, the oxidation number of S in H₂SO₄ is +6.
or, x = +6
Thus, the oxidation number of S in H₂SO₄ is +6.
Oxidation Number of C in CH₃COCH₃:
Let the oxidation number of C in CH₃COCH₃ is x. And the oxidation number of hydrogen and oxygen is +1 and 2 respectively.
∴ 3x + 6(+1) + (2) = 0
or, x = (4/3)
Thus, the oxidation number of C in CH₃COCH₃ is 4/3.
or, x = (4/3)
Thus, the oxidation number of C in CH₃COCH₃ is 4/3.
Oxidation Number of S in Na₂S₂O₃:
Let the oxidation number of S in Na₂S₂O₃ is x.
∴ 2(+1) + 2x + 3(2) = 0
or, x = +2
Thus, the oxidation number of S in Na₂S₂O₃ is +2.
∴ 2(+1) + 2x + 3(2) = 0
or, x = +2
Thus, the oxidation number of S in Na₂S₂O₃ is +2.
Oxidation Number of S in Na₂S₄O₆:
Let the oxidation number of S in Na₂S₄O₆ is x.
∴ 2(+1) + 4x + 6(2) = 0
or, x = +2.5
Thus, the oxidation number of S in Na₂S₄O₆ is +2.5.
∴ 2(+1) + 4x + 6(2) = 0
or, x = +2.5
Thus, the oxidation number of S in Na₂S₄O₆ is +2.5.
Oxidation Number of Cr in [Cr(NH₃)₆]Cl₃:
Let the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is x. NH₃ is neutral thus the oxidation number is zero and the oxidation number of Chlorine is 1.
∴ x + 0 +3(1) = 0
or, x = +3
Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3.
or, x = +3
Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3.
 Problem 2:

Calculate the oxidation number of Iron in [Fe(H_{2}O)_{5}(NO)^{+}]SO_{4}.
 Answer:

H₂O is neutral thus the oxidation number is zero, the oxidation number of (NO)ᐩ is +1 and the oxidation number of SO₄ is 2.

Let the oxidation number of Fe in [Fe(H₂O)₅(NO)ᐩ]SO₄ is x.
∴ x + 0 + (+1) + (2) = 0
or, x  1 = 0
or, x = +1
Thus, the oxidation number of Fe in [Fe(H₂O)₅(NO)ᐩ]SO₄ is +1.
Oxidation Number of an element in a compound is zero:

Some organic compound where the oxidation number of carbon on this compound is zero.
Compound  Formula  Oxidation Number 
Sugar  C_{12}H_{22}O_{11}  0 
Glucose  C_{6}H_{12}O_{6}  0 
Formaldehyde  HCHO  0 

Let, the oxidation number of carbon in Glucose (C12H22O11) is x.
∴ 6x + 12(+1) + 6(2) = 0
or, x = 0
 problem 3:

Calculate the oxidation number of the element marked with blue in the following compounds, (i) K_{2}CrO_{4}, (ii) HOCl, (iii) BaO_{2}, (iv) ClNO, (v) NaNH_{2}, (vi) NaN_{3}, (vii) CH_{2}Cl_{2}, (viii) Ca(OCl)Cl, (ix) Ba(MnO_{4})_{2} (x) CaH_{2}.
 Answer:
Compound  Element  Oxidation Number 
K_{2}CrO_{4}  Cr  +6 
HOCl  Cl  +1 
BaO_{2}  Ba  +2 
ClNO  N  +3 
NaNH_{2}  N  3 
NaN_{3}  N  1/3 
CH_{2}Cl_{2}  C  0 
Ca(OCl)Cl  Cl  +1 
Ba(MnO_{4})_{2}  Mn  +7 
CaH_{2}  Ca  1 
 Problem 4:

What is the Oxidation state of chromium in Cr_{2}O_{5}?
 Answer:

Due to the peroxy linkage oxidation state of Cr in Cr_{2}O_{5} is +6.
Oxidation Number and concept of Oxidation and Reduction:

The oxidation number of CH_{4}, CH_{3}Cl, CH_{2}Cl_{2}, CHCl_{3}, CCl_{4} and CO_{2} are 4, 2, 0, +2, +4 and +4 respectively. In H_{2}, the oxidation number of hydrogen is zero but in H_{2}O it is +1. Similarly, magnesium in the elementary state has a zero oxidation number but in MgCl_{2}, the oxidation number is +2.

From the above, we can define Oxidation and Reduction according to the Oxidation number increase or decrees,
Oxidation:

Oxidation may be defined as a process in which an increase in oxidation number by atoms or ions occurs. A reagent which can increase the oxidation number of an element or ion is called an oxidizing agent.
Reduction:
 The reduction is a process in which a decrease in oxidation number by atoms or ions occurs. A reagent that lowers the oxidation number of an element or ion is called a reducing agent.
Schematic Representation of Oxidation and Reduction According to Electronic Concept and Oxidation Number Concept:

Oxidation and reduction are always found to go hand to hand during a redox reaction. Whenever an element or a compound is oxidized, another element or another compound must be simultaneously reduced.

An oxidant is reduced and simultaneously the reductant is oxidized.
Oxidation number and oxidationreduction 

H2S + Cl2 → 2HCl + S
 Magnesium metal burns in oxygen to produce magnesium oxide: Before the reaction oxidation number of magnesium and oxygen both zero. But after the reaction, the oxidation number of magnesium and oxygen is +2 and 2 respectively. Thus the oxidation number of magnesium is increases and the oxidation number of oxygen is decreased. So magnesium is oxidized and oxygen is reduced.Mg + O2 → 2MgO
 The reaction between Sodium and Chlorine: From the same way, we can explain this reaction. Here oxidation number of Sodium and Chlorine is zero before the reaction and +1 and 1 after the reaction. Thus sodium oxidized and chlorine reduced.Na + Cl2 → 2NaCl
 Problem 5:

Why sulfur dioxide has properties of Oxidation and reduction?
 Answer:

This can be explained by the oxidation state of sulfur. The oxidation numbers of sulfur in the compounds H_{2}S, SO_{2}, and SO_{3} are 2, +4 and +6 respectively. Thus the highest oxidation state of sulfur is +6 and the lowest is 2. The oxidation state of the free sulfur element is 0. In SO_{2}, the oxidation state of sulfur is +4, this oxidation state is middle of 0 and +6.

Thus, the oxidation state can increases from +4 to +6 and decreases from +4 to 0.
SO2 + H2S → 2H2O + 3S