# Application of dipole moment

## Application of dipole moment in chemistry

The application of the dipole moment differentiate between ionic and covalent compound and also provides the framework of inorganic and organic chemistry.

1. When a covalent chemical bond is formed between two identical atoms, the bonding electrons equally sharing by two atoms. Thus the centers of gravity of the two electrons and nuclei therefore coincide.
2. But for two dissimilar atoms, two electrons are not symmetrically disposed. Because each atom has a different attraction for electrons.

When chlorine and bromine combine to form covalent HBr, the electrons forming the covalent bond displaced towards the bromine atom without any separation of the nucleus of an atom.

#### Ionic framework in covalent compounds

Application of dipole moment data uses for the determination of the ionic or covalent character of a heteronuclear diatomic compound.

Let us consider compound HBr having the observed dipole moment = μobs and the bond length l cm.

If we consider HBr is a purely ionic compound the charge on H and Br = 4.8 × 10-10 esu. Thus the dipole moment for this ionic compound

∴ μionic = e × ℓ
= (4.8 × 10-10) ℓ esu cm
where l = bond length

Thus the original dipole moment differs from the calculated dipole moment. This data used to calculate the percentage of ionic character of HBr.

$\therefore&space;Ionic\,&space;character=\left&space;(&space;\frac{\mu&space;_{obs}}{\mu&space;_{ionic}}&space;\right&space;)\times&space;100$

${\color{DarkBlue}&space;=\left&space;(&space;\frac{\mu&space;_{obs}}{4.8\times&space;10^{-10}\times&space;l}&space;\right&space;)\times&space;100}$

where l = bond length of the polar bond.

### Application of induced polarization

The induced polarization used for the determination of molecular radius. The induced electric polarization formula

$P_{i}=\left&space;(&space;\frac{D_{0}-1}{D_{0}+2}&space;\right&space;)\,&space;\frac{M}{\rho&space;}&space;=&space;\frac{4}{3}\,&space;\pi&space;\,&space;N_{0}\,&space;\alpha&space;_{i}$

If  D0 close to unity

$\therefore&space;\left&space;(&space;\frac{D_{0}-1}{3}&space;\right&space;)\times&space;22400=\frac{4}{3}\,&space;\pi&space;\,&space;N_{0}\,&space;r^{3}$

At NTP, M/ρ = molar volume = 22400cc/mole and for spherical molecule, αi= r3.

$r^{3}=\left&space;(&space;\frac{22400}{4\,&space;\pi&space;\,&space;N_{0}}&space;\right&space;)\left&space;(&space;D_{0}-1&space;\right&space;)$$=2.94\times&space;10^{-21}\left&space;(&space;D_{0}-1&space;\right&space;)$

Thus from the known value of capacitance, we can easily determine the radius of the molecules.

### Dipole moment and structure of molecules

The dipole moment application provides the structure and bond angle of different molecules. Thus for mono-atomic noble gases are non-polar because the charge of the constituent atom is distributed symmetrically.

#### Homonuclear diatomic molecules

Nitrogen, oxygen, and chlorine are the homonuclear diatomic molecules with symmetrical charge distributions. Thus the dipole moment of such types of molecules is zero.

#### Heteronuclear diatomic molecules

Hydrogen bromide and hydrogen iodide have non zero values of dipole moment. This indicates the unsymmetrical charge distribution between two bonding atoms.

H+ — I

Due to the difference in electronegativity of the constituent atoms in heteronuclear diatomic molecules always polar.

Thus the electron pair is not equally shared and shifted to the more electronegative atom.

 Hydrogen chloride 1.03 Debye Hydrogen bromide 0.79 Debye Hydrogen iodide 0.38 Debye Hydrogen fluoride 2.00 Debye

#### Calculate the dipole moment of co

The electronegativity difference between carbon and oxygen in CO is very large but the dipole moment of carbon monoxide very low.

This suggested that the charge density in the oxygen atom somehow back-donated to the carbon atom. Thus CO formed a coordinate covalent bond directing towards carbon atom.

### The dipole moment of polyatomic molecules

Carbon dioxide, barium chloride, stannous chloride have zero dipole moment indicating that the molecules have a symmetrical charge distribution between the bond.

Thus in carbon dioxide electric moment of one carbon-oxygen bond cancels the electric moment of the other carbon-oxygen bond.

The electric moment or bond moment associated with the bond arising from the difference of electronegativity.

In compounds, the vectorial addition of the ionic bond moments gives the resultant dipole moment of the compound.

∴ μ2 = m12 + m22 + 2m1m2Cosθ
where m1 and m2 are the bond moments.

Bond moments help to calculate the bond angle of carbon dioxide molecule, where µ = 0 and m1 = m2 = m.

∴ 0 = 2m2(1 + cosθ)
or, θ = 180°

#### The dipole moment of hydrogen sulfide

Water and hydrogen sulfide non-polar because they have non-linear structures. The bond angle can be calculated from the polarity of the molecules.

#### The polarity of the water molecule

Due to the non-linear structure of water molecules, the net dipole moment ≠ 0.

If the dipole moment of water
μ = 1.84 D and bond moment = 1.60 D.

∴ μ2 = 2 m2 (1+ cosθ )
or, (1.84)2 = 2 (1.60)2 (1+ cosθ )
∴ θ = 105°

Thus the contribution of non-bonding electrons towards the total dipole moment included within the bond moment.

#### The polarity of boron trifluoride

Boron trichloride, boron trifluoride are the tetratomic compound having dipole moment zero, indicating that they have regular planar structure.

Their halogen atoms are on a plane at the corner of the equilateral triangle and boron atom at the intersection of the molecules. Thus the bond moment of the above molecules is zero.

#### The polarity of ammonia and phosphine

Other types of the molecule such as ammonia and phosphine are polar, where μ≠0 indicated that the molecule has a pyramidal structure.

Hence three hydrogen atoms on a plane and nitrogen atom at the apex of the pyramid in ammonia and phosphine.

But NF3 shows a very small bond moment although there is a great difference of electronegativity between nitrogen and fluorine atoms and similar structure of NH3.

Thus this low value of µ in NF3 explained by the fact that the resultant bond moment of the three nitrogen – fluorine bonds are acting in the opposite direction to that of the lone pair placed at the nitrogen-atom.

But in NH3, the resultant bond moment is acting in the same direction as that of the lone pair electrons.

#### Penta atomic molecule

Methane, carbon tetrachloride, platinum chloride are examples of Penta-atomic molecules having zero dipole moment.

This suggests that the molecules either regular tetrahedral or square planer structure. But polar molecules of this type have pyramidal structures.

#### The bond polarity of methane

For calculating bond polarity, let us discuss the structure of methane that has regular tetrahedral structure and the angle of each H-C-H = 109°28ˊ.

Thus the group moment of methane 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.

But it can be shown that the group moment of methyl group identical to the bond moment of a carbon hydrogen bond.

Thus the two bond moments cancel each other and the resulting dipole moment of methane is zero.

∴ mCH3 = 3 mCH × Cos(180° -109°28՛)
= 3 mCH Cos 70°32՛
= 3 mCH × (1/3)
So, mCH3 = mCH

Thus µ of methane
= mCH (1 + 3 Cos 109°28՛)
= 0

#### Dipole moment of CH3Cl and CHCl3

μ2 = m12 + m22 + 2 m1m2 Cosθ
But here θ = 0° hence Cosθ = 1

Thus μ2 = m12 + m22 + 2 m1m2
= (m1 + m2)2

∴ μ = (m1 + m2)
= (mCH3 + mCl)
= (1.5 D + 0.4 D)
Thus μCH3Cl = 1.9 D

But for CHCl3

μ = (m1 + m2)
= (mCCl3 + mCH)
= (1.5 D + 0.4 D)
Hence μCHCl3 = 1.9 D

A similar calculation done for the group moment of C2H4, C3H7, C4H9, etc. Thus the bond moment equal to the bond moment of carbon-hydrogen.

But the application of dipole moment gives identical and homologous alcohol and saturated hydrocarbon.