Bonds polarity and dipole moment

The polarity of molecules and bonds

The bond polarity developed when there is an asymmetric change distribution of positive and negative charge among the molecule. But the polarity of a molecule is quantitively expressed by a term called the dipole moment.

Thus evaluation and interpretation of the bonds polarity and dipole moment provide an important tool for the determination of molecular structure.

Thus the application of dipole moment helps to determine the size and shape, spatial arrangement, bond polarity, and residue charge on the atoms of the molecules.

Nonpolar covalent molecule

Homonuclear diatomic molecules are formed between two atoms that possess the same electronegativity. Thus for such cases, chemical bonding electrons are shared equally by two nuclei of the molecule. This type of nonpolar molecules is called non-polar molecules.

H2, N2, O2, Cl2, Br2, I2, etc are examples of such types of molecules.

Thus according to the above rule heteronuclear diatomic molecules are polar due to the difference in their electronegativity. Electronegativity difference is not only the tools to determine the polarity of the molecule, but it also decided by the composition and geometry.

Thus CO2, CS2, BeF2, BCl3, PCl5, methane, benzine, etc are non-polar molecules due to the center of gravity of the positive and negative charge coincide.

Dipole moment polar covalent bond

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 molecule is called polar.

Bonds polarity and net dipole moment of water
Polarity of water

Thus HCl, H2O, NH3, CH3Cl, chlorobenzene are examples of polar covalent molecules.

Definition of the dipole moment

The dipole moment is defined as the product of charge and distance of the separation of the charge.

If +q amount of positive charge separates by -q of negative charge and l is the distance between two centers of the molecule.

Thus  dipole moment, µ = q × l

But for perfectly non-polar compounds shows zero dipole moment and polar ones have positive values. Thus higher the value of the dipole moment of a molecule, higher will be its polarity among the bonds.

Problem
P – F, S – F, Cl – F, and F – F which of the following bonds shows the lowest polarity?

Answer

The electronegativity difference between the two fluorine atoms is zero. Thus the dipole moment of the fluorine among these molecules is zero.

Dipole moment formula

In HCl, due to the greater electronegativity of a chlorine atom, the bonding electron pair shifted towards chlorine atom.

Thus chlorine atom acquires a small negative charge and hydrogen atom acquires a small positive charge. If l is the distance of the charge separation usually taken in bond length.

∴ µ = q × l

The dipole moment is a vector quantity because it has both, magnitude and direction. Thus the direction represented by an arrow pointing towards the negative end and the arrow directly proportional to the magnitude of µ.

CGS and SI unit of the dipole moment

In the CGS system, the charge expressed in esu and the length in cm.

Thus CGS unit of dipole moment = esu cm

Again the charges in the order of 10-10 esu and distance of separation of charge in order of 10-8 cm.

Thus the order of µ
= 10-10 × 10-8
= 10-18 esu cm
This value is known as Debye or simply D.

∴ 1 Debye = 10-18 esu cm

In the SI units, the charge expressed in coulomb and length = meter.

Thus SI unit of the dipole moment
= coulomb × meter (c × m).

What is the value of 1 Debye?

μ = q × l.

Thus the CGS unit of µ
= 4.8 × 10-10 × 10-8 esu cm
= 4.8 D

But the SI unit of µ
1.6 × 10-19 × 10-10 coulombs × meter
= 1.6 × 10-30 C × m

∴ 4.8 Debye = 1.6 ×10-30 coulomb meter

∴1 Debye = 3.336 × 10-30 coulomb meter

The dimension of the dipole moment

Unit of µ = unit of charge × unit of length.
∴ CGS unit of µ = esu × cm

Coulomb’s Law,

    \[ F=\frac{q_{1}\, q_{2}}{D\, r^{2}} \]

Thus (esu)2 = dyne × cm2
= gm cm sec-2 × cm2

    \[ \therefore esu=gm^{\frac{1}{2}}\, sec^{-1}\, cm^{\frac{3}{2}} \]

    \[ unit\, of\, \mu =gm^{\frac{1}{2}}\, sec^{-1}\, cm^{\frac{3}{2}}\times cm \]

    \[ =gm^{\frac{1}{2}}\, sec^{-1}\, cm^{\frac{5}{2}} \]

    \[ \therefore Dimension\, of\, \mu =M^{\frac{1}{2}}L^{\frac{5}{2}}T^{-1} \]