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__Application of Dipole Moment for Molecules: __

__Application of Dipole Moment for Molecules:__

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__(i) For Penta-atomic Molecules(AX₄):
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__(i) For Penta-atomic Molecules(AX₄):__

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 have regular tetrahedral structure and the angle of each H-C-H is 109°28ËŠ.

Structure of CH₄ |

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 with the bond moment of**C-H**(**mC-H**) and so the two moments cancel each other resulting zero dipole moment of the molecule in**CH₄.**Group Moment of CH₄ Molecule |

Thus ,

**m**CH₃ = 3**m**CH Cos(180° -109°28՛) = 3**m**CH Cos 70°32՛ = 3**m**CH × (1/3) =**m**CH####
__Another method to show that resultant moment of CH₄ is zero:__

Resultant Dipole Moment of Methane |

**Î¼ = mCH (1 + 3 Cos 109°28՛) = mCH {1 - (3 ×(1/3)} = 0**

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By similar calculation, it can be shown that mCCl₃ = mCl in CCl₄.

__This idea explains the same value of Î¼ for CH₃Cl and CHCl₃.__

By similar calculation, it can be shown that mCCl₃ = mCl in CCl₄.

__This idea explains the same value of Î¼ for CH₃Cl and CHCl₃.__

Î¼ for CH₃Cl and CHCl₃ |

__For CH₃Cl :__
Thus, Î¼² = m₁² + m₂² + 2 m₁ m₂ CosÎ¸

But here Î¸ = 0° hence CosÎ¸ = 1

∴ Î¼² = m₁² + m₂² + 2 m₁ m₂ = (m₁ + m₂)²

or, Î¼ = (m₁ + m₂) = (mCCl + mCH) = (1.5 D + 0.4 D)

**= 1.9 D**

__For__**:**__CHCl₃__**Î¼**= (m₁ + m₂) = (mCH + mCCl)

= (0.4 D + 1.5 D) =

__1.9 D__
Similar calculation can done for the group moment of

**C₂H₄, C₃H₇, C₄H₉**etc. have the same vale and equal to the bond moment of**C-H**. The identical value of dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.###
__(ii) Hexa Atomic Molecules( AX₅):__

__(ii) Hexa Atomic Molecules( AX₅):__

The molecules of this type are PCl₅, AsCl₅, PF₅ etc have Î¼ =0 indicating that they have pyramidal structure having center of symmetry.

Structure of PF₅ and SF₆ |

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__(iii) Hepta-Atomic Molecules( AX₅):__

__(iii) Hepta-Atomic Molecules( AX₅):__

Hepta-atomic molecules like

**SF₆, XeF₆, WF₆**etc have**Î¼ =0**indicating that these have octahedral symmetrical structure.###
__(iv) Benzene and its Derivatives:__

__(iv) Benzene and its Derivatives:__

Zero dipole moment of the benzene suggested that it has regular hexagonal planer structure and it confirms Kekule’s form.

Regular Planer Hexagonal Structure of Benzene having Centre of Symmetry |

However, if hydrogen atom is substituted by another atom or group, it acquire polar character.

Examples of such derivative of benzene are

**C₆H₅Cl**,**C₆H₅NO₂**,**C₆H₅OH**etc.
When di-substituted benzene are considered it can be shown that o-isomer will have highest value of dipole moment then other two isomers. p-derivative has lowest value while m-derivative has the value between two.

The values can be calculated using the vector addition principle

Î¼ = m₁² + m₂² + 2 m₁m₂ cosÎ¸ provided the group must also be lie on the same plane of the benzene ring.

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__Di-substituted derivative of benzene:__

__Di-substituted derivative of benzene:__

Di-substituted derivative of benzene |

__(a) For o-Isomer:__
Î¼² = m₁² + m₂² + 2 m₁m₂ cosÎ¸

Here, m₁ = m₂ = m and Î¸ = 60°

Thus, Î¼² = 2 m²(1 + cos60°) = 2 m² (1 + 1/2)

**∴**

**Î¼= √3 m**

__(b) For m-Isomer:__
Î¼² = m₁² + m₂² + 2 m₁m₂ cosÎ¸

Here, m₁ = m₂ = m and Î¸ = 120°

Thus, Î¼² = 2 m²(1 + cos120°) = 2 m² (1 - 1/2)

∴

**Î¼ = m**

__(c) For p-Isomer:__
Î¼² = m₁² + m₂² + 2 m₁m₂ cosÎ¸

Here, m₁ = m₂ = m and Î¸ = 180°

Thus, Î¼² = 2 m²(1 + cos180°) = 2 m² (1 - 1)

∴ Î¼ = 0

Thus

**Î¼ of o-isomer〉****Î¼ of m-isomer〉****Î¼ of p-isomer(this has been confirmed for C₆H₄Cl₂ and C₆H₄(NO₂)₂.**
This determination of dipole moment helps to determine the orientation of the groups in the benzene ring.

p - dinitro benzene has Î¼ = 0 but p - dihydroxy benzene Î¼ ≠ 0. Explain.

Through the p-dinitro benzene has Î¼ = 0. p-dihydroxy benzene has dipole moment(Î¼ ≠ 0). This is due to the fact that the two substituted hydroxy group are not on the same plane of the benzene ring but are inclined to the ring.

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__(v) Cis and Trans forms of Geometrical Isomerism:__

__(v) Cis and Trans forms of Geometrical Isomerism:__

Presence of double bond in C - C link restricts the free rotation and geometrical isomerism develops.

Dipole Moment of Cis and Trans Isomers |

The Measurement of dipole moment helps to distinguish the two forms of geometrical isomers.

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__(vi) Change of Dipole Moment with Temperature:__

__(vi) Change of Dipole Moment with Temperature:__

For molecule CH₂Cl - CH₂Cl,

__several conformations are also changed and so dipole moment (Î¼) of the molecule is also changed.__
The dipole moment of o-xylene = 0.693 D. Find the dipole moment of toluene.

For o - xylene the Î¸ = 60° and Î¼ = √3 m. Thus, m = 0.693/√3 = 0.4 D

Again Î¸ = 120° for toluene and dipole moment(Î¼) of toluene = m.

Thus the dipole moment of toluene is 0.4 D.