I am solving some problems involving the order of dipole moments of some compounds. I could solve most of them but the following:

Which has greater dipole moment?

  1. $\ce{CH3F}, \ce{CH_3Cl}$

enter image description here

Since fluorine is more electronegative my assumption was that $\ce{CH3F}$ should have a larger dipole moment.

On the other hand since the angle between the dipole moments is less in 3-nitro phenol than in 4-nitrophenol, my assumption was that 3-nitro phenol has a greater dipole moment. But this contradicts the answer given.


2 Answers 2


The dipole is the vector sum of each atoms's partial charge times its distance from the origin. Lacking this information individual bond dipole values can be used

In the first case it is that of $\ce{CH3}$ vs. the halogen, and then which halogen has the greater electronegativity. Looking up some bond values it seems that a $\ce{C-F}$ bond has the slightly larger dipole ($1.6$ vs $1.5$ Debye ) so the difference is small and not easy to predict as the C-F bond is shorter than the C-Cl.

In the second molecule, geometry can be used as the two groups are the same on each molecule but positioned differently. The para molecule will have the smaller dipole vs. the meta if both groups act in the same direction, either both to withdraw or add electron density with respect to the ring, as then they act to cancel one another to some extent.

If they act in a different direction then the para molecule will have the larger dipole vs the meta. Looking for dipole moments it seems that phenol has a dipole of $1.45$D pointing into the ring and nitrophenol $3.95$D out of the ring so act in different directions and so the para should have the larger dipole.

(To add two vectors of length b and c at an angle $\theta$ the resultant c is found using the cosine rule $c^2 = a^2 + b^2 -2ab\cos(\theta)$ . If the dipoles point in the opposite direction make either a or b negative)


For the second problem $\ce{NO2}$ is able to form resonance structures (being in para position). The resonance forms interupt the shared pi orbitals in the benzene ring (leaving behind two double bonds in the ring) and the nitro group becomes ionized (thus having the greater dipole moment):

Resonance Forms

For the first problem fluorine is more negative, but it lacks valence electrons to withdraw. Chlorine has another valence shell that increases the length of its bond to the methyl group. The increase in bond length leaves a larger space in the sigma bond and more potential to withdraw the electrons farther from the methyl group than the fluorine can.

  • $\begingroup$ Thanks for your answer. But I couldn't get your last para.why is a empty valence shell important.is bond length so high that it is able to counterbalance the electronegativity. $\endgroup$
    – Pink
    Dec 16, 2016 at 5:06
  • 1
    $\begingroup$ chemistry.stackexchange.com/questions/6780/… $\endgroup$
    – user1160
    Dec 16, 2016 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.