Application of Dipole Moment for Molecules

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°28ˊ.
How can we Calculate the dipole Moments of a Molecules?
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.
How can we Calculate the dipole Moments of a Molecules?
Group Moment of CH₄ Molecule
Thus , mCH₃ = 3 mCH Cos(180° -109°28՛) 
= 3 mCH Cos 70°32՛ 
= 3 mCH × (1/3) = mCH

Another method to show that resultant moment of CH₄ is zero:

Resultant Dipole Moment of Methane Molecule
Resultant Dipole Moment of Methane
μ = mCH (1 + 3 Cos 109°28՛) 
mCH {1 - (3 ×(1/3)} = 0
    By similar calculation, it can be shown that mCCl3 = m
    ClCCl4

    This idea explains the same value of μ for CH₃Cl and CHCl₃.

    Applications of Dipole Moment for Pena Atomic Molecules
    μ for CH₃Cl and CHCl₃

    For CH₃Cl :
    Thus, μ²m12 + m22 + 2 m1m2 Cosθ
    But here θ = 0° hence Cosθ = 1
    ∴ μ² = m12m22 + 2 m1m2
    = (m1 + m2
    or, μ = (m1 + m2
    =  (mCCl + mCH
    = (1.5 D + 0.4 D) = 1.9 D
    For CHCl₃:
    μ = (m1 + m2
    =  (mCH + mCCl)
    = (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 dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.

    (ii) Hexa Atomic Molecules( AX₅):

      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. 
      Applications of Dipole Moment for Hexa Atomic Molecules
      Structure of PF₅ and SF₆

      (iii) Hepta-Atomic Molecules( AX₅):

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

        (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.
        Applications of Dipole Moment for Benzene Derivatives
        Regular Planer Hexagonal Structure of
        Benzene having Centre of Symmetry
         
        However, if hydrogen atom is substituted by another atom or group, it acquires 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 the highest value of dipole moment then other two isomers. p-derivative has the lowest value while m-derivative has the value between the two. 
        The values can be calculated using the vector addition principle 
        μ = m12 + m22 + 2 m1m2 cosθ 
        provided the group must also lie on the same plane of the benzene ring.

        Di-substituted derivative of benzene:

        Applications of Dipole Moment for Benzene Derivatives
        Di-substituted derivative of benzene
        (a) For o-Isomer:
        μ² = m12 + m22 + 2 m1m2 cosθ
        Here, m1 = m2 = m and θ = 60°
        Thus, μ² = 2 m²(1 + cos60°) 
        = 2 m² (1 + 1/2) 
        ∴ μ= √3 m
        (b) For m-Isomer:
        μ² = m12 + m22 + 2 m1m2 cosθ
        Here, m1 = m₂ = m and θ = 120°
        Thus, μ² = 2 m²(1 + cos120°) 
        = 2 m² (1 - 1/2)
        ∴ μ = m
        For p-Isomer:
        μ² = m12 + m22 + 2 m1m₂ cosθ
        Here, m1 = m2= m and θ = 180°
        Thus, μ² = 2 m²(1 + cos180°) 
        = 2 m² (1 - 1)
        ∴ μ = 0
        Thus μ of o-isomer〉
        μ of m-isomer〉μ 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.
        • Problem 1:
        p - dinitro benzene has μ = 0 but p - dihydroxy benzene μ ≠ 0. Explain.
        • Answer:
        Through the p-dinitro benzene has μ = 0. p-dihydroxy benzene has dipole moment(μ ≠ 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. 

        (v) Cis and Trans forms of Geometrical Isomerism:

        Presence of double bond in C - C link restricts the free rotation and geometrical isomerism develops.
        Applications of Dipole Moment for Cis and Trans Isomer
        Dipole Moment of Cis and Trans Isomers
        The Measurement of dipole moment helps to distinguish the two forms of geometrical isomers. 

        (vi) Change of Dipole Moment with Temperature:

        For molecule CH₂Cl - CH₂Cl, several conformations are also changed and so dipole moment (μ) of the molecule is also changed.
        • Problem 2:
        The dipole moment of o-xylene = 0.693 D. Finds the dipole moment of toluene.
        • Answer:
        For o - xylene the θ = 60° and μ = √3 m. Thus, m = 0.693/√3 = 0.4 D
        Again θ = 120° for toluene and dipole moment(μ) of toluene = m.
        Thus the dipole moment of toluene is 0.4 D.  

        How to Calculate Dipole Moment of a Compound ? (i) For Penta-atomic Molecules(AX₄), (ii) Hepta-Atomic Molecules( AX₅), (iii) Benzene and its Derivatives.

        Inorganic Chemistry

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