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Showing posts with label DIPOLE MOMENT. Show all posts
Showing posts with label DIPOLE MOMENT. Show all posts

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.

Nov 5, 2018

Application of Dipole Moment for Molecules

How to Calculate Dipole Moment of a Compound?

(i) For Penta-atomic Molecules(AX₄):

The molecules of this type CH₄, CCl₄, PtCl₄ are the examples of having zero dipole moment. This suggests that they are either regular tetrahedral or square planer. But polar molecules of this type have pyramidal structures.
Let us discuss the structure of CH₄ that have regular tetrahedral structure and the angle of each H-C-H is 109°28ˊ.
How can we Calculate the dipole Moments of a Molecules?
Structure of CH₄
The electric moment associated with a group is called group moment and it depends on the arrangement of the bonds in the group and the difference of the electronegativity of the constituent atoms forming the bonds in the group. 
It can be shown that the group moment of CH₃ (mCH₃) is identical with the bond moment of C-H (mC-H) and so the two moments cancel each other resulting zero dipole moment of the molecule in CH₄.
How can we Calculate the dipole Moments of a Molecules?
Group Moment of CH₄ Molecule
Thus , mCH₃ = 3 mCH Cos(180° -109°28՛) 
= 3 mCH Cos 70°32՛ 
= 3 mCH × (1/3) = mCH

Another method to show that resultant moment of CH₄ is zero:

Resultant Dipole Moment of Methane Molecule
Resultant Dipole Moment of Methane
μ = mCH (1 + 3 Cos 109°28՛) 
mCH {1 - (3 ×(1/3)} = 0
By similar calculation, it can be shown that mCCl3 = mClCCl4

This idea explains the same value of μ for CH₃Cl and CHCl₃.

Applications of Dipole Moment for Pena Atomic Molecules
μ for CH₃Cl and CHCl₃
For CH₃Cl :
Thus, μ²m12 + m22 + 2 m1m2 Cosθ
But here θ = 0° hence Cosθ = 1
∴ μ² = m12m22 + 2 m1m2
= (m1 + m2
or, μ = (m1 + m2
=  (mCCl + mCH
= (1.5 D + 0.4 D) = 1.9 D
For CHCl₃:
μ = (m1 + m2
=  (mCH + mCCl)
= (0.4 D + 1.5 D) 
= 1.9 D

Similar calculation can done for the group moment of C₂H₄, C₃H₇, C₄H₉ etc. have the same vale and equal to the bond moment of C-H. The identical value of dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.

(ii) Hexa Atomic Molecules( AX₅):

The molecules of this type are PCl₅, AsCl₅, PF₅ etc have μ =0 indicating that they have pyramidal structure having center of symmetry. 
Applications of Dipole Moment for Hexa Atomic Molecules
Structure of PF₅ and SF₆

(iii) Hepta-Atomic Molecules( AX₅):

Hepta-atomic molecules like SF₆, XeF₆, WF₆ etc have μ =0 indicating that these have octahedral symmetrical structure.

(iv) Benzene and its Derivatives:

Zero dipole moment of the benzene suggested that it has regular hexagonal planer structure and it confirms Kekule’s form.
Applications of Dipole Moment for Benzene Derivatives
Regular Planer Hexagonal Structure of
Benzene having Centre of Symmetry
However, if hydrogen atom is substituted by another atom or group, it acquire polar character.
Examples of such derivative of benzene are C₆H₅Cl, C₆H₅NO₂, C₆H₅OH etc. 
When di-substituted benzene are considered it can be shown that o-isomer will have highest value of dipole moment then other two isomers. p-derivative has lowest value while m-derivative has the value between two. 
The values can be calculated using the vector addition principle 
μ = m12 + m22 + 2 m1m2 cosθ 
provided the group must also be lie on the same plane of the benzene ring.

Di-substituted derivative of benzene:

Applications of Dipole Moment for Benzene Derivatives
Di-substituted derivative of benzene
(a) For o-Isomer:
μ² = m12 + m22 + 2 m1m2 cosθ
Here, m1 = m2 = m and θ = 60°
Thus, μ² = 2 m²(1 + cos60°) 
= 2 m² (1 + 1/2) 
∴ μ= √3 m
(b) For m-Isomer:
μ² = m12 + m22 + 2 m1m2 cosθ
Here, m1 = m₂ = m and θ = 120°
Thus, μ² = 2 m²(1 + cos120°) 
= 2 m² (1 - 1/2)
∴ μ = m
For p-Isomer:
μ² = m12 + m22 + 2 m1m₂ cosθ
Here, m1 = m2= m and θ = 180°
Thus, μ² = 2 m²(1 + cos180°) 
= 2 m² (1 - 1)
∴ μ = 0
Thus μ of o-isomer〉
μ of m-isomer〉μ of p-isomer
(this has been confirmed for C₆H₄Cl₂ and C₆H₄(NO₂)₂.
This determination of dipole moment helps to determine the orientation of the groups in the benzene ring.
p - dinitro benzene has μ = 0 but p - dihydroxy benzene μ ≠ 0. Explain.
Through the p-dinitro benzene has μ = 0. p-dihydroxy benzene has dipole moment(μ ≠ 0). This is due to the fact that the two substituted hydroxy group are not on the same plane of the benzene ring but are inclined to the ring. 

(v) Cis and Trans forms of Geometrical Isomerism:

Presence of double bond in C - C link restricts the free rotation and geometrical isomerism develops.
Applications of Dipole Moment for Cis and Trans Isomer
Dipole Moment of Cis and Trans Isomers
The Measurement of dipole moment helps to distinguish the two forms of geometrical isomers. 

(vi) Change of Dipole Moment with Temperature:

For molecule CH₂Cl - CH₂Cl, several conformations are also changed and so dipole moment (μ) of the molecule is also changed.

The dipole moment of o-xylene = 0.693 D. Find the dipole moment of toluene.
For o - xylene the θ = 60° and μ = √3 m. Thus, m = 0.693/√3 = 0.4 D
Again θ = 120° for toluene and dipole moment(μ) of toluene = m.
Thus the dipole moment of toluene is 0.4 D.  

Dipole Moments


Evaluation and interpretation of dipole moment of covalent molecules provide important tool in the attack of molecular structure. It helps to determine size and shape of the molecules, spatial arrangements of bonds, bonds partial ionic character, residue charge on the atoms of the molecules etc.

Dipole Moment (µ):

The molecules are composed of the partially charged nuclei and negatively charged electrons distributed in space. The structural arrangement of these particles is different in different molecules. When the center of gravity of the positively charge due to coincides with the center of gravity of the negative charge due to electrons, the molecules becomes non polar.
H₂, CO₂, BCl₃, CCl₄, PCl₅, SF₆, 
C₆H₆ etc.
When the center of gravity of the positive charge does not coincide with the center of gravity of the negative charge, polarity arises in the molecules and the molecules are called polar.
HCl, H₂O, NH₃, CH₃Cl, C₆H₅Cl etc.
The polar character of the molecules are quantified a term, called dipole moment (µ). The molecule is neutral and hence if (+q) amount of charge separates at the positive charge center, (- q) will be accumulated at the negative charge center of the molecule. If l is the distance between two centers of the polar molecule, then the dipole moment,
µ = q × l
For the non-polar molecules, l=0 and hence µ=0. Higher the value of µ of a molecule, higher will be its polarity. 
Let us take an example of HCl. Due to greater electronegativity of Cl-atom, the bonding electron pair is shifted towards Cl- atom and it acquire small negative charge (- q) and hydrogen atom acquires small positive charge (+ q). If l is the distance of the charge separation usually taken in bond length, then,
µ = q × l
Dipole moment is a vector quantity and it has both, magnitude and direction. The direction is represented by an arrow pointing towards the negative end.

+ q - q
--l -

The length of the arrow is directly proportional to the magnitude of µ.

Unit of Dipole Moment:

In C.G.S. system, the charge is expressed as esu and the length is cm. 
Thus the unit of µ is esu cm
The charge is the order of 10⁻¹⁰ esu and the distance of separation of charge is in the order of 10⁻⁸ cm. 
Hence the order of µ is 
10⁻¹⁰ × 10⁻⁸ =  10⁻¹⁸ esu cm. 
This magnitude is called one Debye. 
That is, 1 Debye = 10⁻¹⁸ esu cm.
For example, µ of HCl is 1.03 D means, 
µ of HCl is 1.03 х 10⁻¹⁸ esu cm.
In SI system, charge is expresses in coulomb(c) and length is meter(m).
Hence the unit of µ is coulomb × meter (c х m).

How can convert 1 Debye to Coulomb Meter.

The dipole moment in CGS system is
µ = 4.8 × 10⁻¹⁰ × 10⁻⁸ esu cm = 4.8 D
In SI system, 
1.6 ×10⁻¹⁹ × 10⁻¹⁰ coulomb × metre 
= 1.6 × 10⁻³⁰ C×m
Thus, 4.8 Debye = 1.6 ×10⁻³⁰ C×m
or, 1 Debye = (1.6 × 10⁻³⁰)/4.8 C×m
or, 1 Debye = 3.336 × 10⁻³⁰ C×m

Dimension of µ:

Unit of µ = 
Unit of Charge × Unit of Length. 
Thus, the unit of µ 
= esu × cm in CGS system.
But from the Coulomb’s Low, 
F = q1q2/D r²
Thus, (esu)² = dyne × cm²
= gm cm sec⁻² × cm²
Hence, esu = gm½ sec ⁻¹ cm³/₂
Hence the dimension of µ 
M¹/₂ T⁻¹ L⁵/₂

  • Induced Polarization- Clausius Mossotti Equation:

When a non-polar substance is placed between two parallel plates and an electric field is applied, the field tends to attract the negatively charged electrons towards the positive plate and positive charge towards negative plate. 
Under this condition there will be electrical distortion of the molecule and an electric dipole is created. Such distortion in a molecule is called the electric polarization. 
Polarization, however, disappears as soon as the field is withdrawn and the molecule comes back to its original state. 
It is thus the induced polarization (Pi) and the electric dipole is created in the molecule due to the presence of electric field is called induced dipole moment (μi).
The induced dipole moment or simply the induced moment is directly proportional to the strength of the electric field applied(F).
That is, μi ∝ F When F is low, otherwise hyper polarization may occur.
μi = αi F where, αi is proportionality constant, called induced polarisability of the molecule.
αi measures the case with which the electronic configuration of the molecule can be distorted by an applied electric field. It may also be defined as the amount of induced moment in the molecule when unit field strength is applied. The polarisibility has the dimension of the volume (L³).
αi μi/F 
= (esu × cm)/(esu × cm⁻²) 
It can also be shown that, αi = r³ where r = radius of the molecule assuming it to have spherical shape.
For atoms also, distortion occurs when it is placed between the two charged plates. The polarisibility of the atom increases with the atomic size (r), atomic number (Z) and the case of excitation (I.P.). Thus atom behave like dipole and this dipole moment is induced by the applied electric field.
Clausius Mossotti derived from electromagnetic theory, a relation between the polarisibility (αi) and the dielectric constant (D) of the non-polar medium between two plates as,
Clausius Mossotti Equation

Clausius Mossotti Equation for non-polar substance

Where, N₀ denotes Avogadro number, = molar mass and ρ = density of the medium.
The quantity 4/3 π N₀ αi is called molar induced polarization.
It gives the distortion produced in the 1 mole of the substance by a unit electric field. αi is constant for the molecule and independent of temperature. Hence Pi is also constant for the molecule and independent of temperature.
D = dielectric constant of the medium = C/C₀  where C = capacitance of the condenser containing the substance and C₀  is the vacuum. D is dimension less quantity and it is unity for vacuum. For other substance, it is grater then unity. Pi of the substance can be calculated by measuring dielectric constant (D) , density (ρ) and knowing the molar mass (M) of the substance.

  • Orientation Polarization- Debye Modification of Clausius Mossotti Equation:

For polar molecules like CH₃Cl, H₂O, HF etc, molar polarization is not constant but decreases with temperature. 
Thus, Clausius Mossotti Equation fails very badly for polar molecules. The reason for the failure was put forward by P Debye. According to him, when an electric field is applied between two parallel plates containing polar molecules (gaseous state), two effect will occur.
(i) The field will tend to induced polarization in the molecules and the induced polarization in the molecules, 
Pi= (4/3)πN₀αi
(ii)The dipolar molecules will be oriented in the field producing a net dipole moment in the direction of the field.
This polarization is called orientation polarization,
P₀ = (4/3)πN₀α0
Where α0 = orientation polarisibility. 
Debye calculated the value of α0 = μ2/3KT
Considering the two tendency that polar molecules tend to orient in the direction of the applied field (applied) and thermal orientation tends to destroy the alignment of the molecules.
For Non-Polar Molecules
For Non-Polar Molecules
For Polar Molecules
For Polar Molecules
Therefore, for the polar molecules, the total polarization,
Pt = Pi + P0
And the total polarization for polar molecules,
Pt = (D-1)/(D+2) M/ ρ
Dipole moment of the molecules
Debye Equation
However the polar molecules in the substance are fixed and unable to orient in the fixed direction, the orientation polarization is taken zero. This is true for the condensed system where strong inter molecular forces prevent the free rotation of the molecules. 
In that case Pt=Pi
Again if 1/T =0 that is temperature is very high tending to infinity 1/T = 0, Pt=0
And Pt=Pi
This is due to the fact that at high temperature, the polar molecules rotate in such high speed that orientation polarization vanishes.

Application of Debye equation:

This equation can be used to determine the dipole moment of the polar molecules. If Pt is plotted against 1/T , a straight line is obtained and that verify the Debye equation.
The graph of polar and non-polar molecules
The graph of polar and non-polar molecules
The intercept, A = 4/3 π N₀αi from which the induced polarization can be calculated.
The slope, B = 4/3 π N₀(μ²/3k) from which the dipole moment (μ) of the polar molecules can be calculated.
dipole moment (μ) of the polar molecules
Determination of dipole moment
Pt is determined for the substance at different temperatures by determining dielectric constant (D) and density (ρ) at different temperature.