For Pentaatomic 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 HCH is 109°28ˊ.
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 CH (mCH) 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 
μ = m_{CH} (1 + 3 Cos 109°28՛)
= m_{CH} {1  (3 ×(1/3)} = 0 
By similar calculation, it can be shown that m_{CCl3} = m_{ClCCl4}
This idea explains the same value of μ for CH₃Cl and CHCl₃.
μ for CH₃Cl and CHCl₃ 
Thus, μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} Cosθ
But here θ = 0° hence Cosθ = 1
∴ μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2}
= (m_{1} + m_{2})²
or, μ = (m_{1} + m_{2})
= (m_{CCl} + m_{CH}) = (1.5 D + 0.4 D) = 1.9 D 
μ = (m_{1} + m_{2})
= (m_{CH} + m_{CCl})
= (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 CH. 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 pyramidal structure having center of symmetry.
Structure of PF₅ and SF₆ 
(iii) HeptaAtomic Molecules( AX₅):
Heptaatomic molecules like SF₆, XeF₆, WF₆ etc have μ =0 indicating that these have 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.
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 disubstituted benzene are considered it can be shown that oisomer will have highest value of dipole moment then other two isomers. pderivative has lowest value while mderivative has the value between two.
The values can be calculated using the vector addition principle
μ = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ
provided the group must also be lie on the same plane of the benzene ring.
provided the group must also be lie on the same plane of the benzene ring.
Disubstituted derivative of benzene:
Disubstituted derivative of benzene 
μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ
Here, m_{1} = m_{2} = m and θ = 60°
Thus, μ² = 2 m²(1 + cos60°)
= 2 m² (1 + 1/2)
∴ μ= √3 m

μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m_{2} cosθ
Here, m_{1} = m₂ = m and θ = 120°
Thus, μ² = 2 m²(1 + cos120°)
= 2 m² (1  1/2)
∴ μ = m

μ² = m_{1}^{2} + m_{2}^{2} + 2 m_{1}m₂ cosθ
Here, m_{1} = m_{2}= m and θ = 180°
Thus, μ² = 2 m²(1 + cos180°)
= 2 m² (1  1)
∴ μ = 0

Thus μ of oisomer〉
μ of misomer〉μ of pisomer
(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 pdinitro benzene has μ = 0. pdihydroxy 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.
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 oxylene = 0.693 D. Find the dipole moment of toluene.
 Answer:
Again θ = 120° for toluene and dipole moment(μ) of toluene = m.
Thus the dipole moment of toluene is 0.4 D.