Arrange the following sequence in order of increasing dipole moment.
Toluene $ (1) $
m-dichlorobenzene $ (2) $
o- dichlorobenzene $ (3) $
p- dichlorobenzene $ (4) $
(A) $ 1 < 4 < 2 < 3 $
(B) $ 4 < 1 < 2 < 3 $
(C) $ 4 < 1 < 3 < 2 $
(D) $ 4 < 2 < 1 < 3 $
Answer
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Hint :Dipole moment is developed between two atoms with a huge difference in their electronegativities. The bond that takes place between two atoms is covalent bond. If the electronegativities are more the dipole moments are more too but it also depends on symmetry.
Complete Step By Step Answer:
Let us first understand why dipole moment is is generated,
Let us consider a bond between carbon and a halogen $ C - X $ . A covalent bond is generated when there is a huge difference between the electronegativities of two atoms. In the given example $ C - X $ the electronegativity of carbon is $ 2.6 $ and let us consider the halogen as chlorine who’s electronegativity is $ 3.5 $ . As we can see the difference between carbon and chlorine is that a very polar bond is generated.
Now, when we consider the given option toluene, there is a negligible difference between the electronegativities hence a very small amount of dipole moment .i.e. $ 0.6D $ is generated (the bond between $ CH{}_3 $ and benzene ring is $ sp{}_2\& sp{}_3 $ ).
Considering, o-dichlorobenzene, m- dichlorobenzene and p- dichlorobenzene have derivatives as $ \alpha = 60^\circ ,120^\circ \& 180^\circ $ and hence have resultant vector as zero dipole moment in para derivative due to its symmetrical structure. The dipole moment of meta dichlorobenzene is more than toluene. Ortho and meta dichlorobenzene have higher dipole moments than toluene because of its high electronegativity of chlorine and $ CH{}_3 $ . Ortho-dichlorobenzene has a lower bond angle than the meta isomer resulting in higher dipole moment
Hence the correct option is B, p-dichlorobenzeneWith the dipole moments of $ 0 < 0.36 < 1.73 < 2.54 $ .
Note :
We must know the factors to give the correct option. Those factors are structure, symmetry, electronegativity, bond formation, and dipole moment. The unit for dipole moment is debye and it is written as D.
Complete Step By Step Answer:
Let us first understand why dipole moment is is generated,
Let us consider a bond between carbon and a halogen $ C - X $ . A covalent bond is generated when there is a huge difference between the electronegativities of two atoms. In the given example $ C - X $ the electronegativity of carbon is $ 2.6 $ and let us consider the halogen as chlorine who’s electronegativity is $ 3.5 $ . As we can see the difference between carbon and chlorine is that a very polar bond is generated.
Now, when we consider the given option toluene, there is a negligible difference between the electronegativities hence a very small amount of dipole moment .i.e. $ 0.6D $ is generated (the bond between $ CH{}_3 $ and benzene ring is $ sp{}_2\& sp{}_3 $ ).
Considering, o-dichlorobenzene, m- dichlorobenzene and p- dichlorobenzene have derivatives as $ \alpha = 60^\circ ,120^\circ \& 180^\circ $ and hence have resultant vector as zero dipole moment in para derivative due to its symmetrical structure. The dipole moment of meta dichlorobenzene is more than toluene. Ortho and meta dichlorobenzene have higher dipole moments than toluene because of its high electronegativity of chlorine and $ CH{}_3 $ . Ortho-dichlorobenzene has a lower bond angle than the meta isomer resulting in higher dipole moment
Hence the correct option is B, p-dichlorobenzene
Note :
We must know the factors to give the correct option. Those factors are structure, symmetry, electronegativity, bond formation, and dipole moment. The unit for dipole moment is debye and it is written as D.
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