Oxidation Number

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. 
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.
2H2 + Cl2  2HCl
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 
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+2 + 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 super oxides 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 + O-2 
(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). 
Oxidation number of P in Ba(H2PO2)2 is - (a)+3, (b)+2, (c) +1, (d) -1.
(c) +1
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 -1.
Similar way, 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.
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
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 = +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, 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
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, 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, 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, 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
Thus, the oxidation number of Cr in [Cr(NH3)6]Cl3 is +3.
Calculate the oxidation number of Iron in [Fe(H2O)5(NO)+]SO4.
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)+]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)+]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 C12H22O11 0
Glucose C6H12O6 0
Formaldehyde HCHO 0
Let, the oxidation number of carbon in Glucose (C12H22O11) is x
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0
Calculate the oxidation number of the element marked with blue 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.
Compound Element Oxidation Number
K2CrO4 Cr +6
HOCl Cl +1
BaO2 Ba +2
ClNO N +3
NaNH2 N -3
NaN3 N -1/3
CH2Cl2 C 0
Ca(OCl)Cl Cl +1
Ba(MnO4)2 Mn +7
CaH2 Ca -1
What is the Oxidation state of chromium in Cr2O5?
Due to peroxy linkage oxidation state of Cr in Cr2O5 is +6.

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 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 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:

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.
Oxidation Number and Concept of Oxidation and Reduction.
Schematic Representation of Oxidation and Reduction.
| Oxidation
H2S + Cl2 2HCl + S
Reductant Oxidant Oxidant Reductant
| Reduction
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.
0 0 +2  -2
Mg + O2 2MgO
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.
0 0 +1  -1
Na + Cl2 2NaCl
Why sulphur dioxide has properties of Oxidation and reduction?
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.
| Oxidation Number Decreases
Reduction
SO2 + H2S 2H2O + 3S
| Oxidation Number Increases
Oxidation



Oxidation Number and Concept of Oxidation and Reduction: Method of finding out the Oxidation Number, Oxidation Number of an Element in a compound, Oxidation Number and concept of Oxidation and Reduction

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