## What is dipole moment in chemistry?

**Dipole moment** in chemistry expressed the polar character or polarity of the molecules. The product of charge and distance of separation of atoms in chemical bond defines the term dipole moment of molecules. If +q amount positive charge separated by −q amount of negative charge by the bond distance l, then dipole moment (μ) of polar molecule = q × l.

Perfectly non-polar molecules like hydrogen, oxygen, and nitrogen have zero dipole moment. The polar molecules like water, ammonia, and methane have positive values. It is a tool used to calculate the percentage ionic character, electric polarization, and residual charge on the atoms of the molecules.

### What are polar molecules?

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. These molecules are called polar molecules. Hydrogen chloride, water, methyl chloride, and benzyl chloride are examples of polar molecules.

## Unit of dipole moment

In the CGS system, the charge is expressed in esu and bond length in cm. Therefore, the unit of dipole moment is esu cm. The charge is of the order of 10^{−10} esu and the distance of separation is the order of 10^{−8} cm. Hence the order of μ is 10^{−18} esu cm. This magnitude is called 1 Debye. Therefore, 1 Debye = 10^{−18} esu cm.

In the SI system, the charge is expressed in coulomb and the length is meter. Hence the unit of dipole moment in the SI system is the coulomb meter.

### Debye to coulomb meter

In the CGS system,

μ = 4.8 × 10^{−10} × 10^{−8} esu cm

= 4.8 Debye.

In the SI system,

μ = 1.6 × 10^{−19} × 10^{−10} coulomb m

= 16 × 10^{−30} coulomb m.

Therefore, 4.8 Debye = 16 × 10^{−30} coulomb m. Form the above formula, 1 Debye = 3.336 × 10^{−30} coulomb m.

## Application of dipole moment

It is used for the calculation of the percentage ionic character, bond angle, electric polarization, and residual charge on the atoms in the molecules. It also helps to determine the size or shape of molecules and the arrangements of chemical bonds in the molecules.

## Percentage ionic character

Dipole moment data is used to calculate the percentage ionic character of covalent or ionic heteronuclear diatomic molecules. Let us consider a molecule AB having the observed dipole moment = μ_{obs} and the bond length l cm.

- If the shared pair lies at the midpoint of the atoms, the bond would be a purely covalent bond, and the percentage ionic character is zero.
- But if the bond is 100 percent ionic and B is more electronegative than A. Therefore, A carries a unit positive charge, and B carries a unit negative charge.

### Percentage ionic character formula

In that case, the μ of AB would be, μ_{ionic} = e × l = 4.8 × 10^{−18} esu cm. But the μ of AB is neither zero nor μ_{ionic}.

### Percentage ionic character of HCl

If we consider HCl as a purely ionic compound the charge on hydrogen and chlorine = 4.8 × 10^{−10} esu and bond length = 1.27 × 10^{−8} cm.

The dipole moment of HCl,

μ_{ionic} = e × l

= 4.8 × 10^{−10} × 1.27 × 10^{−8} esu cm

The calculated or observed dipole moment = 1.03 Debye = 1.03 × 10^{−18} esu cm.

Hence the percentage ionic character of HCl = (µ_{obs}/µ_{ionic}) × 100 = (1.03 × 10^{−18}/4.8 × 10^{−10} × 1.27 × 10^{−8}) × 100 = 16.89.

### Electric polarization formula

In learning chemistry, the induced electric polarization formula is used to calculate the radius of the molecules. The induced electric polarization formula of the molecules is,

At NTP, M/ρ = molar volume or density = 22400cc/mole and for spherical molecule, α_{i}= r^{3}. Therefore, r^{3} = (22400/4πN_{0})(D_{0} −1) = 2.94 × 10^{−21} (D_{0} −1). Therefore, with the known value of capacitance, we can easily determine the radius of the molecule.

## Dipole moment examples

It is used to define the structure, bond angle, bond energy, and polarity of different molecules in chemistry. Mono-atomic noble gases and benzene are non-polar because the charge of the constituent atom is distributed symmetrically.

### Homonuclear diatomic molecules examples

Homonuclear diatomic molecules like nitrogen, oxygen, and chlorine have zero dipole moment due to the symmetrical charge distributions and similar electronegativity and ionization energy.

### Heteronuclear diatomic molecules examples

Hydrogen bromide and hydrogen iodide have non-zero dipole moments indicating the unsymmetrical charge distribution between two bonding atoms in the molecules.

Due to the difference in electronegativity of the constituent atoms in heteronuclear diatomic molecules are always polar. Hence the electron pair is not equally shared in the hydridized orbital and shifted to the more electronegative atom. Therefore, μ_{HCl} = 1.03 Debye, μ_{HBr} = 0.79 Debye, μ_{HI} = 0.38 Debye, μ_{HF} = 2.00 Debye.

### Dipole moment of CO

The electronegativity difference between carbon and oxygen in CO is very large but the dipole moment of carbon monoxide is very low. This suggested that the charge density in the oxygen atom somehow back-donated to the carbon atom. Hence CO formed a coordinate covalent bond directed towards the carbon atom to decrease polarity.

## How to calculate dipole moment?

Carbon dioxide (CO_{2}), beryllium chloride (BeCl_{2}), and stannous chloride (SnCl_{2}) have zero dipole moment indicating that the molecules have symmetrical linear structures. In CO_{2}, one carbon-oxygen bond cancels the bond moment of the other carbon-oxygen bond.

### What is bond moment?

The electric moment associated with the bond arising from the difference in electronegativity is called the bond moment (m). In a molecule, the vectorial addition of the bond moments is used for the calculation of the μ.

Hence, μ^{2} = m_{1}^{2} + m_{2}^{2} + 2m_{1}m_{2}Cosθ

where m_{1} and m_{2} are the bond moments projecting at the angle of θ. Bond moment and dipole moment calculation help to find the bond angle of CO_{2}, H_{2}O, H_{2}S, SO_{2}, etc.

For CO_{2}, µ = 0 and m_{1} = m_{2 }= m.

Therefore, 0 = 2m^{2}(1 + cosθ)

or θ = 180°

For H_{2}O, H_{2}S, and SO_{2} molecules, θ < 180°. Therefore, these molecules are non-polar because they have non-linear structures.

Molecule |
Formula |
Dipole moment (Debye) |
Bond angle |

Carbon dioxide | CO_{2} |
0 | 180° |

Water | H_{2}O |
1.84 | 105° |

Hydrogen sulfide | H_{2}S |
0.94 | 97° |

Beryllium chloride | BeCl_{2} |
0 | 180° |

Sulfur dioxide | SO_{2} |
1.61 | 119° |

### Dipole moment of water molecule

For water (H_{2}O), dipole moment (μ_{water}) = 1.84 Debye and bond moment (m_{OH}) = 1.60 Debye.

Therefore, (1.84)^{2} = 2(1.60)^{2}(1 + cosθ)

Form the above formula θ = 105°. Therefore, the contribution of non-bonding electrons toward the total dipole moment is included within the bond moment of water.

### Dipole moment of BF_{3} and BCl_{3}

Boron trichloride (BCl_{3}), and boron trifluoride (BF_{3}) are the tetratomic compound having dipole moment zero, indicating that they have a regular planar structure.

From the above picture, three halogen atoms are on a plane at the corner of the equilateral triangle and a boron atom at the intersection of the medians. Therefore, the μ_{net} of the BCl_{3} and BF_{3} is zero.

### Dipole moment of NH_{3} and PH_{3}

Other types of the molecule such as ammonia (NH_{3}) and phosphine (PH_{3}) are polar, where μ≠0 indicated that the molecule has a pyramidal structure. Hence three hydrogen atoms are on the plane and nitrogen or phosphorus atoms at the apex of the pyramid in the NH_{3} or PH_{3} molecule.

### Dipole moment of NF_{3}

NF_{3} shows a very small bond moment although there is a great difference in electronegativity and electron affinity between nitrogen and fluorine atoms and a similar structure of NH_{3}.

- The low value of µ in NF
_{3}is explained by the fact that the resultant bond moment of the three nitrogen-fluorine bonds is acting in the opposite direction to that of the lone pair placed at the nitrogen atom. - But in NH
_{3}, the resultant bond moment is acting in the same direction as that of the lone pair electrons.

## Penta atomic molecule

Methane (CH_{4}), carbon tetrachloride (CCl_{4}), and platinum chloride (PtCl_{4}) are examples of penta atomic molecules having zero dipole moment. This suggests that they are either regular tetrahedral or square planer structures. But the polar molecules of this type have pyramidal structures.

### Dipole moment of methane

For the calculation of the dipole moment of methane, let us discuss the structure of methane. In methane molecule, the valence shell electronic configuration of central carbon atom 2s^{2} 2p^{2}. Hence the carbon atom in methane sp^{3} hybridized to form a regular tetrahedral structure with the angle of each H-C-H = 109°28ˊ.

The electric moment associated with a group is called a group moment. It depends on the arrangement of the bonds in the groups. The difference in the electronegativity of the constituent atoms forming the bonds in the group.

It can be shown that the group moment of the methyl group (m_{CH3}) is identical to the bond moment of a carbon hydrogen bonding (m_{CH}).

From the structure and bond moment,

m_{CH3} = 3 m_{CH} × Cos(180° -109°28՛)

= 3 m_{CH} Cos 70°32՛

= 3 m_{CH} × (1/3)

Therefore, m_{CH3} = m_{CH}

Therefore, two bond moments cancel each other to show zero dipole moment in the CH_{4} molecule. It can be also calculate by another methods, µ_{methane} = m_{CH} (1 + 3 Cos 109°28՛) = 0.

## Dipole moment measurement

### Dipole moment of CCl_{4}

By similar calculation, it can be shown that m_{CCl3} = m_{CCl} = m in chloroform (CCl_{4}). Therefore, the net dipole moment for CCl_{4} molecule (μ_{chloroform}) = 2m^{2}(1 + Cosθ) = 0, where Cosθ = Cos180° = −1.

### Dipole moment of chloromethane (CH_{3}Cl)

We know, μ^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} Cosθ. But for dichloromethane (CH_{3}Cl), θ = 0° hence Cosθ = 1. Therefore, μ_{chlorometane} = (m_{1} + m_{2}) = (1.5 D + 0.4 D) = 1.9 D

### Dipole moment of CHCl_{3}

Similarly, for trichloromethane (CHCl_{3}), θ = 0° hence Cosθ = 1. Therefore, μ_{CHCl3} = (m_{1} + m_{2}) = (1.5 D + 0.4 D) = 1.9 D.

### Dipole moment of hydrocarbons

A similar calculation is done for μ of hydrocarbon like ethylene, propylene, butylene, etc, and alcohol in organic chemistry. The bond moment of m_{OH} = 1.6, m_{C-O} = 0.7, m_{CH3} = m_{CH} = 0.4 Debye. Therefore, the resultant dipole moment of methyl alcohol = 1.56 while the observed value = 1.65 debye.