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The covalent bond form between two or more having different electronegativity the atom having higher electronegativity will draw the bonded electron pair more towards itself resulting in a partial charge separation. The distribution of the electron cloud in the bond does not remain uniform and shifts towards the more eletronegative one. Such bonds are called polar covalent bonds. and molecules of the type H – X having two polar ends (positive and negative) are known as polar molecules. For example the bond formed between hydrogen and chlorine or between hydrogen and oxygen in water.

The extent of polar character or the degree of polarity in a compound is determined by the term dipole moment. 


(1) Polarity of any polar covalent bond or molecule is measured in terms of dipole moment(μ).
(2) For measurement of extent of polarity, Pauling introduced the concept of dipole moment .
"The product of positive or negative charge (q) and the distance (d) between two poles is called  dipole moment". Here
                            [Dipole moment= magnitude of charge x distance]
                                                  [μ = q x d] 
                                   Where μ= dipole moment
                                              q= Charge in esu unit 
                                              d= distance between two charges
Dipole moment is usually expressed in Debye (D)

Let charge on one electron in esu unit = 4.8×10-10 esu {(e=1.6×10-19 × 3.0×10-9) (1C=3.0×10-9)}
Consider a standard dipole of length (d) 1A° or 10-8 cm and charge (q) 4.8×10-10 esu.
                   μ= 4.8×10-10×10-8 cm
                  μ= 4.8×10-18 esu cm
                  μ=4.8 D (1×10-18 esu cm = 1debye or 1D)
Let charge on one electron in S.I. unit =1.6×10-19 coulomb's or C 
Consider a standard dipole of length (d) 1A° or 10-10 m and charge (q) 1.6×10-19
                   μ= 1.6×10-19 ×10-10 Cm
                   μ= 1.6×10-29 C m
On comparing value of dipole moment in both units  
             4.8 D=1.6×10-29 Cm
                 1D= 3.3×10-30 Cm
(3) Dipole moment is a vector quantity i.e. it has both magnitude as well as direction.
(5) In the diatomic molecule dipole moment (μ) depends upon difference of electronegativity i.e. dipole moment (μ) electronegativity order of dipole moment (μ)  H–F > H–Cl > H–Br > H–I. and dipole moment (μ) = 0 for  H–H, F–F, Cl–Cl, Br–Br, O=O
(6) For poly atomic molecule dipole moment (μ) depends on the vector sum of dipole moments of all the covalent bonds.

(7) For PCl5 and SF6, etc. dipole moment (μ) = 0 due to their regular geometry.

(8) Benzene, naphthalene, biphenyl have dipole moment (μ) = 0 due to planar structure.
(9) If the vector sum is zero, than compound is non-polar compound or symmetrical compound (and it is not essential that individual dipole moments (μ) of every bond should be zero).

 ILLUSTRATIVE EXAMPLE(A): BX3, CCl4, SiCl4, CH4, CO2, CS2, PCl5, SiH4 etc.  In these examples the bond B–F, C–Cl, C–H, C–O, P–Cl etc. are polar even though compounds are non-polar.

(10) Dipole moment of H2O is 1.85 D which is resultant dipole moment (μ) of two O–H bonds. Dipole moment (μ) of H2O is more than dipole moment (μ) of H2S because electronegativity oxygen is higher than sulphur.
(11) Angular structure of molecule have greater dipole moment.

ILLUSTRATIVE EXAMPLE (1): CO2 has got dipole moment of zero why?

SOLUTION: The structure of CO2 is this is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is the negative end. Here two equal dipoles acting in opposite direction cancel each other and therefore the dipole moment is zero.
ILLUSTRATIVE EXAMPLE (2): Dipole moment of CCl4 is zero while that of CHCl3 is non zero.
SOLUTION: Both CCl4 & CHCl3 have tetrahedral structure but CCl4 is symmetrical while CHCl3 is non-symmetrical
Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero. But in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry.

ILLUSTRATIVE EXAMPLE (3): Compare the dipole moment of H2O and F2O.
SOLUTION: Let’s draw the structure of both two compounds and then analyses it.
In both H2O and F2O the structure is quite the same. In H2O as O is more electronegative than hydrogen so the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and dipole moment of H2O is high. In case of F2O the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole.
ILLUSTRATIVE EXAMPLE (4): Compare the dipole moment of NH3 and NF3.?
Hence μ NH3 is greater than μ NF
ILLUSTRATIVE EXAMPLE (5): but -2- ene. It exists in two forms Cis and Trans
SOLUTION: The Trans isomer is symmetrical with the 2-methyl groups in anti position. So the bond dipoles the two Me– C bonds acting in opposition cancel each other result in a zero dipole. Whereas in cis isomer the dipoles do not cancel each other resulting in a net dipole.

ILLUSTRATIVE EXAMPLE (6): Compare the dipole moment of Cis 1,2 trichloroethylene and trans 1,2 trichloroethylene. 
SOLUTION: In the Trans compound the C-Cl bond dipoles are equal and at the same time acting in opposition cancel each other while C is compound the dipoles do not cancel each other resulting in a higher value. Generally all Trans compounds have a lower dipole moment corresponding to C is isomer, when both the substituent’s attached to carbon atom 
are either electron releasing or electron withdrawing.


To locate position of substituents in aromatic compounds
( A)  If same substituents are present in the symmetrical position dipole moment (u) of benzene ring compounds will be zero.

(B) As angle between substituent’s decrease, value of dipole moment (u )increas.



ILLUSTRATIVE EXAMPLE (7):Out of Meta and ortho isomers 4-methylnitrobenzene which is having greater dipole moment?
SOLUTION: Para isomer is having highest dipole moment since two groups attached to benzene ring have dipole moment directed in the same direction thereby they reinforce one another in this case.
ILLUSTRATIVE EXAMPLE (8): The resultant dipole moment of water is 1.85 D ignoring the effects of lone pair. Calculate, the bond moment of each O-H bond (given that bond angle in H2O = 104°, cos 104° = –0.25)
                                              R2 = P2 + Q2 + 2PQ cos(theta)



ILLUSTRATIVE EXAMPLE (9): Calculate the % of ionic character of a bond having length = 0.92 Å and 1.91 D as its observed dipole moment ?.
SOLUTION:   To calculate μ considering 100% ionic bond
                      = 4.8 × 10–10 × 0.92 × 10–8esu cm
                      = 4.8 × 0.83 × 10–18 esu cm = 4.416 D
                         ∴ % ionic character = 1.91/4.416× 100 = 43.25                     
ILLUSTRATIVE EXAMPLE (10): Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as its observed dipole moment?
SOLUTION: To calculate μ considering 100% ionic bond
                      = 4.8 × 10–10 × 0.83 × 10–8esu cm
                      = 4.8 × 0.83 × 10–18 esu cm = 3.984 D
                         ∴ % ionic character = 1.82/3.984 × 100 = 45.68


Q 1: Which of the following has been arranged in order of decreasing dipole moment?
Q 2: Which has maximum dipole moment?

Ans (A) Due to the symmetrical structure dipole moment of (C) & (D) are zero & (A) having maximum dipole moment.


Q. (1) the dipole moment of HBr is 7.95 Debye and the inter molecular separation is 1.94 x10-10 m Find the % ionic character in HBr molecule.
Q. (2) HBr has dipole moment 2.6x10-30 C-m. If the ionic character of the bond is 11.5%, calculate the inter atomic spacing.
Q.(3) Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate  percentage ionic character in LiF molecule Li--F bond length is 156 pm.
Q. (4) A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 Å, what percentage of an electronic charge exists on each atom.
 Ans:  (1).    85 %, (2)   1.4 Å, (3).     84.5% ,  (4).  25%   

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