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Nov 4, 2018

Oxidation Number

Oxidation Number 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.
For Examples, formation of water from hydrogen and oxygen, can not be covered by electronic concept since water is not an ionic compound.
2H₂ + O₂ 2H₂O
It may recall classically we could still say that hydrogen is oxidised to H₂O. In the same seance burning of magnesium in oxygen is considered oxidation.
Similarly hydrogen and chlorine react react to form a covalent molecule hydrogen chloride.
H₂ + Cl₂ HCl
The Above reaction hydrogen is oxidised or chlorine is reduced but the resulting compound is covalent one, thus the reaction can not be covered by the electronic concept. 
In order to cover such reactions also under oxidation-reduction 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 arbitrary 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 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 same electronegativity are involves for bonding. 

2. Except of metal hydrides 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 
For Examples,
NaH Na⁺ + H⁻ Here oxidation number of H is (-1)
HCl H⁺ + Cl⁻ Here oxidation number of H is (+1)
3. The oxidation number of metal is positive. 
For Examples, 
 CuO Cu⁺⁺ + O⁻ ⁻ 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 super oxides the oxidation number of oxygen is -1 and -1/2 respectively. 
In fluorine monoxide(F₂O) oxygen has an oxidation number +2 because fluorine is more electronegative than oxygen. 
For Examples, 
CuO Cu⁺⁺ + O⁻ ⁻ Here oxidation number of O is (-2). 
In H₂O, Oxidation Number of Oxygen is (-2), but in H₂O₂, Oxidation Number of H is (+1) and Oxidation Number of Oxygen is (-1). In BaO₂, Oxidation Number of Ba is (+2) and thus the Oxidation Number of Oxygen is (-1). In Na₂O, Oxidation Number of Na is (+1) and Oxidation Number of oxygen is (-2). 
Problem:
Oxidation number of P in Ba(H₂PO₂)₂ is - (a)+3, (b)+2, (c) +1, (d) -1.
Answer: (c)+1
Solution:
Oxidation number of Ba is +2, oxidation number of hydrogen is +1 and 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. 
For examples, 
NaCl Na⁺ + Cl⁻
The charge and oxidation number of Na⁺ is +1 and the charge and oxidation number of Cl⁻ is -1Similar way MgBr₂, MgBr₂ Mg⁺⁺ + 2Br⁻ Here the charge of Mg⁺⁺ 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.
For Examples, 
In HCl oxidation number of hydrogen is +1 and oxidation number of chlorine is -1. And the sum of these = (+1) + (-1) =0.

Oxidation Number of an Element in a compound: 

1. 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=0; or, x=+7 
Thus, the oxidation number of Mn in KMnO₄ is +7. 
2. 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-8=-2; or, x=+6 
Thus, the oxidation number of Mn in MnO₄⁻² is +6. 
3. Oxidation Number of Cr in Cr₂O₇⁻² : 
Let the oxidation number of Cr in Cr₂O₇⁻² is x 
∴ 2x+7(-2)=-2; or, 2x-14=-2; or, x=+6 
Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6 .
4. 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=0; or, x=+6 
Thus, the oxidation number of S in H₂SO₄ is +6. 
5. 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, 3x+4=0; or, x=–(4/3) 
Thus, the oxidation number of C in CH₃COCH₃ is -(4/3) 
6. Oxidation Number of S in Na₂S₂O₃:
Let the oxidation number of S in Na₂S₂O₃ is x. 
∴ 2(+1)+2x+3(-2)=0; or, 2+2x-6=0; or, x=+2 
Thus, the oxidation number of S in Na₂S₂O₃ is +2 
7. Oxidation Number of S in Na₂S₄O₆:
Let the oxidation number of S in Na₂S₄O₆ is x. 
∴ 2(+1)+4x+6(-2)=0; or, 2+4x-12=0; or, x=+2.5 
Thus, the oxidation number of S in Na₂S₄O₆ is +2.5 
8. 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=0; or, x=+3 
Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3
Problem:
Calculate the oxidation number of Iron in [Fe(H₂O)₅(NO)⁺]SO₄.
Answer:
H₂O is neutral thus the oxidation number is zero, oxidation number of (NO)⁺ is +1 and the oxidation number of SO₄ is -2.
Let the oxidation number of Fe in [Fe(H₂O)₅(NO)⁺]⁺² 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)⁺]⁺² 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. As for Example, In Sugar (C₁₂H₂₂O₁₁), Glucose (C₆H₁₂O₆), Formaldehyde (HCHO) etc. the oxidation number of carbon=
Let, the oxidation number of carbon in Glucose (C₆H₁₂O₆) is x. 
∴ 6x+12(+1)+6(-2)=0; or, x=0 
Problem:
Calculate the oxidation number of the element marked with line in the following compounds, (i) K₂CrO₄, (ii) HOCl, (iii) BaO₂, (iv) ClNO, (v) NaNH₂, (vi) NaN₃, (vii) CH₂Cl₂, (viii) Ca(OCl)Cl, (ix) Ba(MnO₄)₂ (x) CaH₂. 
Answer: 
(i) Cr +6, (ii) Cl +1, (iii) Ba +2, (iv) N +3, (v) N -3, (vi) N -1/3, (vii) C  0, (viii) Cl +1 Cl -1, (ix) Mn +7 (x) H -1.

Oxidation Number and concept of Oxidation and Reduction:

The oxidation number of CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ and CO₂ are -4, -2, 0, +2, +4 and +4 respectively. In H₂, the oxidation number of hydrogen is zero but in H₂O it is +1. Similarly magnesium in elementary state has a zero oxidation number but in MgCl₂, the oxidation number is +2.
From the above discussion we see that, the oxidation of same element increases or decreases and define oxidation and reduction.
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 oxidising agent. 
Reduction:
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:
Schematic Representation of Oxidation and Reduction
Schematic Representation of Oxidation and Reduction
Oxidation and reduction are always found to go hand to hand during a radox reaction. Whenever an element or a compound is oxidised, another element or another compound must be simultaneously reduced.
An oxidant is reduced and simultaneously the reductant is oxidised.
Representation of oxidant and reductant
Representation of oxidant and reductant
Examples:
1.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 oxidation number of oxygen is decreases. So magnesium is oxidised and oxygen is reduced.
Representation of Oxidation and Reduction
Representation of Oxidation and Reduction
2. Reaction between Sodium and Chlorine:
From 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 oxidised and chlorine reduced.
Reaction between Sodium and Chlorine
Representation of Oxidation and Reduction

The same substance acting as an oxidising and a reducing agent:
Problem:
Why sulphur dioxide has properties of Oxidation and reduction?
Answer:
This can be explain by the oxidation state of sulphur. The oxidation numbers of sulphur in the compounds H₂S, SO₂, and SO₃ are -2, +4 and +6 respectively. Thus the highest oxidation state of sulphur is +6 and lowest is -2. The oxidation state of free sulphur element is 0. In SO₂, the oxidation state of sulphur 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.