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Jan 1, 2019

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.
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)⁺]⁺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

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.

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
H -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:

What is the Oxidation Number? How can we find out the oxidation number of an element in a Compound?
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.
Oxidation Number and Oxidation Number of an Elements in a Compound.
Representation of oxidant and reductant

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.
How can we define oxidation and Reduction According to Oxidation Number Concept?
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.
0
0
+1
-1
2Na
+
Cl₂
2
Na
Cl

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.



Dec 31, 2018

Application of Dipole Moment

Application of Dipole Moment in Chemistry: 

Different applications are taken one after another.

1.Determination of Partial Ionic Character and Residual Charge on the Constituent Atoms in a Molecule:


Let us consider a molecule A - B having the dipole moment μobs and the bond length l cm. if the shared electron pair lies at the midpoint of the atoms, the bond will be purely covalent and the present of ionic character is zero. 
But if the bond is 100% ionic, and B is more electronegative then A, B will carry unit negative charge and A uni-positive charged respectively. In that case the dipole moment of the molecule would be,
μionic = e × 
= (4.8 × 10⁻¹⁰ esu cm
But the dipole moment of AB is neither zero or nor μionic .
How to Calculate Percent Ionic Character with Dipole Moment and Bond Length
Determination of Partial Ionic Character and Residual Charge

The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. 
calculate the (a) charge on the constituent atom and 
(b) the % of the ionic Character of HCl.

Given, μobs = 1.03 Debye 
= 1.03 × 10⁻¹⁸ esu cm and length ℓ 
= 1.27 × 10 ⁻⁸ cm.
(a). Hence the charge on the constituent atom(q),
μobs/ = (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻⁸)
= 0.8 × 10⁻¹⁰ esu 
(b) The % of the ionic Character of the molecule,
(μobs/μionic ) × 100 =(1.03×10⁻¹⁸)/(4.8×10⁻¹⁰×1.27×10⁻⁸)×100 
= 16.89%

  • Determination of molecular radius from induced polarization (Pi) :


The induced polarization (Pi)
= (4/3) π N0 αi 
But when the substance is in the gaseous state, 
Pi = {(D0-1)/(D0+2)} M/ρ 
(for the covalent substance) 
The value of D൦ is close to unity under this condition, 
Hence, {(D0 - 1)/3} × 22400 
= ( 4/3 ) π N0 r³
At NTP, M/ρ = molar volume = 22400cc/mole and αi= r³ taking the spherical shape of the molecule. 
or, r³ = (22400/4πN0 ) (D0 - 1) 
2.94 ×10⁻²¹ (D൦ - 1) 
Hence, radius of the molecule r can be determined by measuring D൦ of the substance at NTP.

  • Determination of molecular structure :


(i) Mono-atomic Molecules (A) : 

The mono-atomic inert gases are non-polar and it indicates the symmetrical charge distribution in molecule. 

(ii) Di-atomic Molecules (AX) :

The homonucler diatomic molecules are largely non-polar such as N₂, O₂, Cl₂ indicating that there are symmetrical charge distributions. The bonding electron pair is equally shared by the two bonding atoms. However, Br₂, I₂ have non zero vale of dipole moment and this indicates the unsymmetrical charge distribution,
I+ I-
Compounds Dipole Moment Structure
H2 0 H H
Cl2 0 Cl Cl
Heteronucler diatomic molecules are always polar due to difference of electronegativity of the constituent atoms. The example are , HCl, HBr, HF etc. this indicates that electron pair is not equally shared and shifted to the more electronegative atom.
δ+ δ-
HCl 1.03 D H Cl
δ+ δ-
HBr 0.79 D H Br
δ+ δ-
HI 0.38 D H I
δ+ δ-
HF 2.00 D H F

The Difference between electronegativity of Carbon and Oxygen Is large but the dipole moments of carbon monoxide is very low - Why?

However, in CO, there are large difference of electronegativity between C and O but the molecule is very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom. This explain by forming a coordinate covalent bond directing towards C-atom.
How to Calculate Percent Ionic Character with Dipole Moment and Bond Length
Formation of Coordinate Covalent Bond in Carbon Monoxide

(iii) Tri-atomic Molecules (AX₂):

The molecules, like CO₂, BeCl₂, BeF₂, SnCl₂ etc have zero dipole moment indicating that the molecules have symmetrical linear structure. 
For example, CO₂ has structure, The electric moment of one C - O bond ( known as bond moment ) cancel the electric moment of the other C - O bond.
δ- δ+ δ-
O = C = O
δ- δ+ δ-
Cl = Be = Cl
The electric moment associated with the bond arising from difference of electronegativity is called the bond moment(m). In molecules, the vectorial addition of the bond moments (m) gives the resultant dipole moment(μ) of the molecule.
μ² = m12 + m22 + 2m1m2Cosθ
Where, m1 and m2 are the bond moments.
These helps to calculate the bond angle (θ) in the molecule. Thus in the carbon dioxide molecule, μ = 0 and m1 = m2 
hence, 0 = 2m²(1 + cosθ) 
or, θ = 180° 
that is the molecule is linear.
For Non-Linear Structure
Another type of molecules such as, H₂O, H₂S, SO₂᠌᠌ etc. have μ ≠ 0 indicating that they have non linear structure. The bond angle can be calculated from the bond moments of the molecules.

H₂O molecule has a dipole moment-Explain. Does it invalidate a linear structure ?

Bond Moment and Dipole Moment
Dipole Moment of H₂O and H₂S
For Water (H₂O) , μ = 1.84 D 
and mOH = 1.60 D 
Thus, μ² = 2 m² (1+ cosθ ) 
or, (1.84)² = 2 (1.60)² (1+ cosθ ) 
or, θ = 105° 
The contribution of non-bonding electrons towards the total dipole moment is included within the bond moment.

The bond angle in H₂S is 97° and dipole moment = 0.95 D . Find the bond moment of the S - H link. ( Given, Cos97° = -0.122)

We have, μ = 0.95 D and θ = 97° . 
From the equation, 
μ² = 2 m² (1+ cosθ ) 
Putting the value we have, 
(0.95)² = 2 m² (1 + cos97° ) 
here m = mS-H 
or, 0.9025 = 2 m² (1-0.122) 
or, m² = 0.9025/ (2 × 0.878) 
or, m² = 0.5139 
or, m = 0.72 
Thus the bond moment of the S - H link is 0.72 D

(iv) Tetra Atomic Molecules (AX₃) : 

The molecules like BCl₃, BF₃ etc have dipole moment zero indicating that they have regular planar structure. There halogen atoms are on a plane at the corner of the equilateral triangle and B atom is at the intersection of the molecules.
Bond Moment and Dipole Moment
Regular Planar Structure Of BF 

Show that the bond moment vectors of BF₃ molecule adds up to zero.

Bond Moment and Dipole Moment
Resultant Dipole Moment of BF
While other type of the molecule such as NH₃, PH₃, AsH₃ are polar (μ≠0) indicating that that the molecule have a pyramidal structure in which the three H-atoms are on the plane and N-atom at the apex of the pyramid in NH₃. 
But NF₃ has very small dipole moment though there is great difference of electronegativity between N and F atoms and similar structure of NH₃.
Bond Moment and Dipole Moment
Resultant Dipole Moment of NH and BF
Low value of μ of NF₃ is explain by the fact that resultant bond moment of the three N - F bonds is acting in opposite direction to that of the lone pair placed at the N-atom. But in NH₃, the resultant bond moment is acting in the same direction as that of the lone pair electrons.