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The usual explanation for the molecular dipole moment of NF$_3$ being smaller than that of NH$_3$, despite the N-F dipole being stronger than the N-H dipole, is that influence of the lone electron pair on nitrogen is to oppose the net N-F$_3$ dipole, while enhancing that of N-H$_3$. See, for example, this good example.

What bothers me about this explanation is that when the N-F dipole is calculated (experimentally, I think), the effect of the electron pair on N is already factored in. Likewise for the calculation for N-H.

Putting the question in terms of the language of the linked article: "[T]he lone pair's negative concentration will augment the polarity contribution from the polar bonds while for the NF3, the lone pair's dipole would subtract from the dipole contributed by the NF bonds."

If the lone pair on N subtracts from the NF bond in the molecule it must have had the same effect in the original diatomic dipole calculation. Likewise for NH.

If this is true, assuming identical N-F and N-H bond angles in NF$_3$ and NH$_3$, it seems that the vector calculation should still give a stronger dipole moment for NF$_3.$ [Please regard this as a question!]

One thing that would lessen the molecular NF$_3$ dipole moment is a smaller NF bond angle (compared to NH), which increases the lateral component of the dipole at the expense of the axial, making the (axial) resultant smaller. I think fluorine is larger than hydrogen and maybe this would tend to push the fluorine atoms out laterally, decreasing the resultant axial dipole. However I do not see this as an explanation in texts so I am guessing it is wrong.

My confusion is that the vector algebra doesn't seem to support the argument but there is probably something I am not taking into consideration.

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  • $\begingroup$ Do you have a table of bond dipoles as a reference? $\endgroup$ – Buck Thorn Aug 29 at 8:22
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You should probably be cautious in general applying the concept of bond dipoles to predict molecular dipole moments, and it does seem ad hoc if not downright flimflam to invoke a dipole to the nitrogen lone pair to predict dipole moments in the examples you present (particularly in the case of nitrogen trifluoride).

Now if you read the wikipedia page on nitrogen trifluoride, it explains that

NF3 is slightly soluble in water without undergoing chemical reaction. It is nonbasic with a low dipole moment of 0.2340 D. By contrast, ammonia is basic and highly polar (1.47 D).[9] This difference arises from the fluorine atoms acting as electron withdrawing groups, attracting essentially all of the lone pair electrons on the nitrogen atom. NF3 is a potent yet sluggish oxidizer.

The low basicity of $\ce{NF3}$ supports the argument that electron density associated with unpaired electrons on N is withdrawn by the fluorine atoms. So it would appear that in the case of $\ce{NF3}$ there might not be a significant dipole moment between the N nucleus and an electron pair (or the charge opposite the three fluorines). Instead the electron density (relative to ammonia as a reference) is polarized toward the fluorines. This makes the application of a method of linear summation of bond dipoles somewhat sketchy.

Continuing along similar sketchy lines, the website to which you provided a link argues that the difference in molecular dipoles cannot be explained simply in terms of the order of electronegativity (F>N>H). But note that electronegativities are computed from dissociation energies, electron affinities, etc, and not from partial charges or dipole moments, and the relation between charges or dipole moments and energies is bound to be highly non-linear, making quantitative prediction of dipole moments based on electronegativities difficult.

What seems necessary is to perform some ab initio electronic structure calculations.

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  • $\begingroup$ OK but if F withdraws all the electron density there would still be a large net dipole moment. Yet it is smaller than that of $NH_3.$ Maybe it is true that calculations are missing and if so that's a good answer. +1 Flimflam? $\endgroup$ – daniel Aug 28 at 11:31
  • $\begingroup$ @daniel flimflam is a good word :-) even if my usage is not necessarily accurate as the original user perhaps did not intend to deceive. I'll modify my answer, to make a little more clear what I mean, when I get a chance. $\endgroup$ – Buck Thorn Aug 29 at 8:09

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