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.
IONIC NATURE IN COVALENT BOND:
(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)
CALCULATION OF DIPOLE MOMENT:
IN C.G.S UNIT:
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)
IN S.I. UNIT:
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.
ILLUSTRATIVE EXAMPLE(B):
(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.?
SOLUTION:
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.
are either electron releasing or electron withdrawing.
DIPOLE MOMENT IN AROMATIC RING SYSTEM:
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.
CALCULATION AS BELOW:
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)
SOLUTION:
R2 = P2 + Q2
+ 2PQ cos(theta)
SOME IMPORTANT ORDERS OF DIPOLE MOMENTS (u)
TO CALCULATE % IONIC CHARACTER:
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
MULTIPLE CHOICE QUESTIONS(MCQ)
Q 1: Which of the following has been
arranged in order of decreasing dipole moment?
Ans-(A)
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.
EXERCISE:
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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