A different application of dipole moment is taken one after another.

### Determination of partial ionic character

- Let us consider a molecule AB having the dipole moment μobs and the bond length l cm. If the shared electron pair lies at the midpoint of the atoms, the bond will be purely covalent and the present of ionic character is zero.

- But if the bond is ionic, and B is more electronegative then A, B will carry a unit negative charge and a uni-positive charged respectively. In that case, the dipole moment of the molecule would be,

μ

= (4.8 × 10⁻¹⁰) ℓ esu cm

_{ionic}= e × ℓ= (4.8 × 10⁻¹⁰) ℓ esu cm

- But the dipole moment of AB is neither zero or nor μ

_{ionic}. Thus we can calculate partial ionic character by the application of dipole moment.

Ionic character |

### Induced polarization

The induced polarization (Pi)

= (4/3) π N₀ αi

But when the substance is in the gaseous state,

Pi = {(D₀ - 1)/(D₀ + 2)} M/ρ

(for the covalent substance)

The value of D൦ is close to unity under this condition,

Hence, {(D0 - 1)/3} × 22400 = ( 4/3 ) π N₀ r³

At NTP, M/ρ = molar volume = 22400cc/mole and

αi= r³ taking the spherical shape of the molecule

or, r³ = (22400/4πN₀ ) (D₀ - 1)

= 2.94 ×10⁻²¹ (D₀ - 1)

= (4/3) π N₀ αi

But when the substance is in the gaseous state,

Pi = {(D₀ - 1)/(D₀ + 2)} M/ρ

(for the covalent substance)

The value of D൦ is close to unity under this condition,

Hence, {(D0 - 1)/3} × 22400 = ( 4/3 ) π N₀ r³

At NTP, M/ρ = molar volume = 22400cc/mole and

αi= r³ taking the spherical shape of the molecule

or, r³ = (22400/4πN₀ ) (D₀ - 1)

= 2.94 ×10⁻²¹ (D₀ - 1)

- Hence, the radius of the molecule r can be determined by measuring D൦ of the substance at NTP.

### Molecular structure form dipole moment

#### Mono-atomic molecules

- The mono-atomic inert gases are non-polar, and it indicates the symmetrical charge distribution in the molecule.

#### Di-atomic molecules

- The homonuclear diatomic molecules are largely non-polar such as N₂, O₂, Cl₂ indicating that there are symmetrical charge distributions. The bonding electron pair is equally shared by the two bonding atoms.

- However, Br₂, I₂ have non zero values of dipole moment and this indicates the unsymmetrical charge distribution,

H⁺ㄧI⁻

- Heteronuclear diatomic molecules are always polar due to the difference in electronegativity of the constituent atoms. The example is, HCl, HBr, HF, etc. this indicates that the electron pair is not equally shared and shifted to the more electronegative atom.

Dipole moment of HCl, HBr, HI, HF |

#### Tri-atomic molecules

- The molecules, like CO₂, BeCl₂, BeF₂, SnCl₂, etc have zero dipole moment indicating that the molecules have a symmetrical linear structure.

- For example, CO₂ has structure, The electric moment of one C - O bond ( known as bond moment ) cancels the electric moment of the other CO bond.

Dipole moment of CO₂ and BeCl₂ |

- The electric moment associated with the bond arising from the difference of electronegativity is called the bond moment(m). In molecules, the vectorial addition of the bond moments (m) gives the resultant dipole moment(μ) of the molecule.

μ

^{2}= m_{1}^{2}+ m_{2}^{2}+ 2m_{1}m_{2}CosθWhere m₁ and m₂ are the bond moments.

- These help to calculate the bond angle (θ) in the molecule. Thus in the carbon dioxide molecule, μ = 0 and m₁ = m₂

Hence, 0 = 2m²(1 + cosθ) or, θ = 180° that is the molecule is linear.

#### Non-linear structure

- Another type of molecules such as H₂O, H₂S, SO₂᠌᠌, etc. have μ ≠ 0 indicating that they have non-linear structure. The bond angle can be calculated from the bond moments of the molecules.

#### Tetra atomic molecules

- The molecules like BCl₃, BF₃, etc have 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 B atom is at the intersection of the molecules.

Application of dipole moment of BF₃ |

- While other types of the molecule such as NH₃, PH₃, AsH₃ are polar (μ≠0) indicating that that the molecule has a pyramidal structure in which the three H-atoms are on the plane and N-atom at the apex of the pyramid in NH₃. ;

- But NF₃ has a very small dipole moment though there is a great difference of electronegativity between N and F atoms and similar structure of NH₃.

- A low value of μ of NF₃ is explained by the fact that the resultant bond moment of the three N - F bonds are acting in the opposite direction to that of the lone pair placed at the N-atom. But in NH₃, the resultant bond moment is acting in the same direction as that of the lone pair electrons.

Dipole Moment of NH₃ and BF₃ |

#### Penta-atomic molecules

- The molecules of this type CH₄, CCl₄, PtCl₄ are the examples of having zero dipole moment. This suggests that they are either regular tetrahedral or square planer. But polar molecules of this type have pyramidal structures.

- Let us discuss the structure of CH₄ that has regular tetrahedral structure and the angle of each H-C-H is 109°28ˊ.

CH₄ molecule |

- The electric moment associated with a group is called group moment and it 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.

- It can be shown that the group moment of CH₃ (mCH₃) is identical to the bond moment of C-H (mC-H) and so the two moments cancel each other resulting in zero dipole moment of the molecule in CH₄.

Thus , mCH₃ = 3 mCH Cos(180° -109°28՛)

= 3 mCH Cos 70°32՛

= 3 mCH × (1/3)

= mCH

= 3 mCH Cos 70°32՛

= 3 mCH × (1/3)

= mCH

#### Dipole moment of CH₄

Thus the resultant dipole moment of CH₄

= mCH (1 + 3 Cos 109°28՛)

= mCH {1 - (3 ×(1/3)}

= 0

= mCH (1 + 3 Cos 109°28՛)

= mCH {1 - (3 ×(1/3)}

= 0

#### Dipole moment of CCl₄, CHCl₃, and CH₃Cl

CCl₄, CHCl₃, and CH₃Cl molecules |

μ² = m12 + m22 + 2 m1m2 Cosθ

But here θ = 0° hence Cosθ = 1

∴ μ² = m12 + m22 + 2 m1m2= (m1 + m2)²

or, μ = (m1 + m2)

= (mCCl + mCH)

= (1.5 D + 0.4 D)

= 1.9 D

But here θ = 0° hence Cosθ = 1

∴ μ² = m12 + m22 + 2 m1m2= (m1 + m2)²

or, μ = (m1 + m2)

= (mCCl + mCH)

= (1.5 D + 0.4 D)

= 1.9 D

#### Dipole moment of CHCl₃ molecules

μ = (m1 + m2)= (mCH + mCCl)

= (0.4 D + 1.5 D)

= 1.9 D

= (0.4 D + 1.5 D)

= 1.9 D

- A similar calculation can be done for the group moment of C₂H₄, C₃H₇, C₄H₉, etc. have the same value and equal to the bond moment of C-H.

- The identical value of the dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.

#### Hexa atomic molecules

- The molecules of this type are PCl₅, AsCl₅, PF₅, etc have μ =0 indicating that they have a pyramidal structure having the center of symmetry.

#### Hepta-atomic molecules

- Hepta-atomic molecules like SF₆, XeF₆, WF₆, etc have μ =0 indicating that these have an octahedral symmetrical structure.

### Questions answers

Question- The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. Calculate the (a) charge on the constituent atom and (b) the % of the ionic Character of HCl.

Answer

(a) Charge on the constituent atom(q), = 0.8 × 10⁻¹⁰ esu

(b) Percentage of the ionic character of HCl, = 16.89%

Question(b) Percentage of the ionic character of HCl, = 16.89%

- The difference between the electronegativity of carbon and oxygen is large but the dipole moments of carbon monoxide are very low - why?

Answer

- However, in CO, there is a large difference in electronegativity between C and O but the molecule is the very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom.

- This explains by forming a coordinate covalent bond directing towards C-atom.

- H₂O molecule has a dipole moment-explain. Does it invalidate a linear structure?

Dipole Moment of H₂O and H₂S |

For Water (H₂O), μ = 1.84 D and mOH = 1.60 D

Thus, μ² = 2 m² (1+ cosθ )

or, (1.84)² = 2 (1.60)² (1+ cosθ )

or, θ = 105°

Thus, μ² = 2 m² (1+ cosθ )

or, (1.84)² = 2 (1.60)² (1+ cosθ )

or, θ = 105°

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

- The bond angle in H₂S is 97° and dipole moment = 0.95 D. Find the bond moment of the S - H link. ( Given, Cos97° = -0.122)

We have, μ = 0.95 D and θ = 97°.

From the equation, μ² = 2 m² (1+ cosθ )

Putting the value we have, (0.95)² = 2 m² (1 + cos97° )

Here m = mS-Hor, 0.9025 = 2 m² (1-0.122)

or, m² = 0.9025/ (2 × 0.878)

or, m² = 0.5139

or, m = 0.72

Thus the bond moment of the S - H link is 0.72 D

From the equation, μ² = 2 m² (1+ cosθ )

Putting the value we have, (0.95)² = 2 m² (1 + cos97° )

Here m = mS-Hor, 0.9025 = 2 m² (1-0.122)

or, m² = 0.9025/ (2 × 0.878)

or, m² = 0.5139

or, m = 0.72

Thus the bond moment of the S - H link is 0.72 D