I was asked in one question that which of all given molecules have a permanent dipole moment?

And $\ce{XeF6}$ was given as a correct answer quoting that it is a distorted octahedral, and hence the lone pair results in the molecule having a dipole moment. But since we know that the lone pair goes at each face of the octahedral one by one, the net effect should cancel out. Hence, is it correct to say that the molecule has a PERMANENT dipole?

I am having similar doubt in the case of the very fundamental ammonia example. Since it undergoes inversion do we say that it has a permanent dipole?

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

  • 2
    $\begingroup$ In the equilibrium (ie: lowest energy) situations, both NH3 and XeF6 have a permanent dipole moment. Since the molecule is very likely to be around its equilibrium geometry at any given time, then the general characteristics of the compound resemble those of the equilibrium situation. So, despite undergoing inversion via an apolar intermediate, ammonia is not an apolar molecule, and so isn't XeF6 $\endgroup$
    – user32223
    Commented Jun 6, 2020 at 7:16
  • 3
    $\begingroup$ Forget inversion. Molecules wiggle, bounce, and rotate all the time; how can we talk about the dipole moment? Now it points this way, and the next second it points that way. Is there any meaning in it at all? Well, yes, there is. $\endgroup$ Commented Jun 6, 2020 at 10:20
  • 1
    $\begingroup$ You might look at chemistry.stackexchange.com/questions/104468/… but I'm not quite sure if that is the question that is being asked (I think more than one this is being conflated here). For me for a molecule to have a permanent electric dipole means the expectation value of the electric dipole operator with the molecule in its equilibrium geometry is non-zero. The effect on the measured dipole due to motion is a separate issue. $\endgroup$
    – Ian Bush
    Commented Jun 6, 2020 at 13:40
  • $\begingroup$ @Ian Bush: Second part of the question is definitely related to: chemistry.stackexchange.com/questions/104468/…. $\endgroup$ Commented Jun 6, 2020 at 14:06

3 Answers 3


Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1):

Crystal Structures of XeF6

The abstract of Ref.1 tells them all:

According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential scanning calorimetry, $\ce{XeF6}$ exists in at least six different modifications. Three of them are formed at temperatures above room temperature, one exists at room temperature, while two have been found at low temperatures. In the high temperature modifications $\ce{XeF6}$ forms a non-symmetric tetramer, better described as a cyclic trimer with a weakly associated monomer. The normal temperature modification is the previously described cubic phase IV, having disordered tetrameric and hexameric units. The low temperature modifications are regular tetramers. Only in presence of $\ce{HF}$ symmetric dimers are formed.

The tetramer $\bf{C}$ (ball and stick version of $\bf{A}$) represent the general version of all modifications, a cyclic trimer with a weakly associated monomer. The symmetric dimer $\bf{D}$ is the structure of $\ce{XeF6}$ in the presence of $\ce{HF}$ (highlighted by $\color{green}{\text{green}}$ ovals). Nonetheless, all of these modifications display distorted octahedral structure for individual $\ce{XeF6}$ molecule as represented by ball & stick 3D structure in the central box. None of $\ce{F-Xe-F}$ angles are equal to $90^\circ$, which is the normal angle of central atom with two adjacent ligands in regular octahedron. As a consequence, $\ce{XeF6}$ has net permanent dipole moment since all six $\ce{Xe^\delta+-F^\delta-}$ bonds are polar.

For your question about ammonia, look at Ivan Neretin's comment and here where there is an accepted answer.


  1. Sevim Hoyer, Thomas Emmler, Konrad Seppelt, “The structure of xenon hexafluoride in the solid state,” Journal of Fluorine Chemistry 2006, 127(10), 1415-1422 (https://doi.org/10.1016/j.jfluchem.2006.04.014).

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

Addressing your second question, permanent dipole moment experimentally means that a molecule will be deflected by an electric field. Consider water as an example, open your kitchen faucet slowly, so that a think thin stream is steadily flowing. Charge a comb and see its deflection. Such molecules have a permanent dipole moment.

If you had cyclohexane or heptane (a molecule without a permanent dipole moment) nothing will happen, i.e., their streams will not be deflected by an electric field.

Always focus on basic understanding first before being bogged down by complex cases of ammonia inversion or invoking operators for electric dipole moments etc.

enter image description here

  • $\begingroup$ Ok, thanks for the information. So would a liquid ammonia stream get deflected in an electric field or would the rapid inversions cancel out dipole moment of each other? $\endgroup$ Commented Jun 6, 2020 at 16:55
  • $\begingroup$ Ammonia has a dipole moment which is permanent. Whether it is an inverted state or or normal state, the individual bond dipole moments do not cancel. Hence liquid ammonia will be deflected by an electric field. $\endgroup$
    – ACR
    Commented Jun 6, 2020 at 17:07

The distorted octahedral structure of $\ce{XeF6}$ produces a permanent dipole. The key word is distorted structure, which results from the $\ce{Xe}$ lone pair).

3D structre od XeF6

If a structure is distorted, this results in a (lowering) breaking of symmetry. If a molecule is unsymmetrical with strong polarising groups, there will be a local permanent dipole moment associated with it. In a general sense, you can think of bonds as vectors.

Now in your $\ce{XeF6}$ example:

  1. Consider the direction of the bonds as vectors.

  2. Then calculated (or visually consider) the sum of those vectors.

  3. If the sum of all vectors is not equal to 0, then you have a dipole moment.

Now consider an perfectly octahedral molecule. You will find that the vector sum is now equal to zero, and therefore the molecule does not contain a permanent dipole.

Now for your $\ce{NH3}$ example. $\ce{NH3}$ has a distorted tetrahedral shape, where the bonds are forced away from the lone pair. The mistake is to assume bonds and lone-pairs are the same. The electron negative charge within the lone pair is held closer to the centre of the atom within the lone-pair orbital, and so repels the electrons in the bonds to a greater extent.

To contrast $\ce{:NH3}$ (here, $\ce{:}$ represent lone pair) with $\ce{CH4}$ (which doesn't have a permanent dipole). All bonded vectors in $\ce{CH4}$ are equal (same atom-bond type) and are evenly distributed.

  • 6
    $\begingroup$ Good as it is, your answer is from a different question. $\endgroup$ Commented Jun 6, 2020 at 10:17

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.