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Nov 1, 2018

Application of Dipole Moment

Application of Dipole Moment Measurement:

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 .
ionic character of molecules
Determination of Partial Ionic Character and Residual Charge
Problem:
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.
Answer:
Given, μobs = 1.03 Debye = 1.03 × 10⁻¹⁸ esu cm and length ℓ = 1.27 × 10 ⁻⁸cm.
(a). Hence the charge on the constituent atom(q),
  • = μobs/ℓ = (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻) = 0.8 × 10⁻¹⁰ esu 
(b) The % of the ionic Character of the molecule,
(μobsionic ) ×100 
=(1.03 × 10⁻¹⁸)/(4.8 ×10⁻¹⁰×1.27 × 10⁻×100
= 16.89%

Determination of molecular radius from induced polarization (Pi):

The induced polarization (Pi) = (4/3)  π N꧐ αi
But when the substance is in the gaseous state,
Pi = {(D൦-1)/(D൦+2)}   M/ρ (for the covalent substance)
The value of D൦ is close to unity under this condition,
Hence, {(D൦ - 1)/3}  × 22400 = (  4/3 ) π N൦ r³At NTP, M/ρ = molar volume = 22400cc/moleAnd αi taking the spherical shape of the molecule.
or, r³ = (22400/4πN൦ ) (D൦ - 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⁻
 dipole moment of hydrogen and chlorine molecule
Dipole Moment Of Hydrogen And Chlorine
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.
Dipole Moment of HCl, HBr, HI, HF
Dipole Moment of HCl, HBr, HI, HF
Problem:
The Difference between electronegativity of Carbon and Oxygen Is large but the dipole moments of carbon monoxide is very low - Why?
Answer:
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.
Formation of Coordinate Covalent Bond in Carbon Monoxide
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.
Structure of carbon dioxide and berriliym chloride
Dipole Moment Representation of CO₂, BeCl₂
The electric moment associated with the bond arising from the difference of electronegativity is called the bond moment(m). In molecules, the vectorial addition of the bond moments (m) gives the resultant dipole moment(μof the molecule.
That is, μ² = m₁² + m₂² + 2m₁m₂Cosθ
Where, m and m 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.
Problem:
H₂O molecule has a dipole moment-Explain. Does it invalidate a linear structure ?
Answer:


Dipole Moment of H₂O and H₂S
Dipole Moment of H₂O and H₂S
Example:
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.
Problem : 
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)
Solution:
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.
Tetra Atomic Molecules (AX₃)
Regular Planar Structure Of BF 
Problem:
Show that the bond moment vectors of BF₃ molecule adds up to zero. 
Answer:
Tetra Atomic Molecules (AX₃)
Resultant Dipole Moment of BF
Let μ is the resultant moment of the molecule along the direction represent on the above picture. 
Then, μ = m(1 + cos120° + cos240°) = (1 - 1/2 -1/2) = 0
Hence net moment of the BF₃ molecule is zero. 
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.
Tetra Atomic Molecules (AX₃)
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.