It is established the dipole moment is a result of multiplication of the magnitude of charges (Q) and the distance between them (r). What I understand is that when an electron and a proton get closer (bond length decreases), polarity decreases hence the dipole decreases. When they move further away from each other(bond length increases), the polarity increases and hence the dipole moment increases. However, in case of halides (H-X), let's take for example HF and HI, HF has a bigger dipole moment than HI, it is stated that the bond length of HF is smaller than HI. Why do I find this contradictory? Can you help me with What I am missing here?
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asked Sep 19, 2017 at 19:15
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$\begingroup$ First part of your first statement? $\endgroup$– MithoronCommented Sep 19, 2017 at 19:44
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$\begingroup$ If you relate and compare two properties (here bond lenght and dipole moment) quantitatively, you have to look if other relevant properties (here the effective charge!) are changing. Which they are. $\endgroup$– KarlCommented Sep 19, 2017 at 19:55
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$\begingroup$ related chemistry.stackexchange.com/questions/6780/… $\endgroup$– MithoronCommented Sep 20, 2017 at 0:18
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2 Answers
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$\dagger$ For those who might want to point out possible vibrational excitation or some other subtle (or not-so-subtle) effect: don't. Let's keep it simple.
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Strictly speaking, you can neither decrease nor increase the bond length a slightest bit.$\dagger$ All molecules of a given compound are equal. It is not like you may grab one with tiny pincers of some sort, stretch it a little and see what happens.
You may, however, consider a molecule of a different compound; indeed, that's what you do in your question. But that's not just a different bond length; it is different atoms with different properties, and hence different everything.
Most importantly for the question at hand, iodine has much lower electronegativity than fluorine, so the effective charges on $\rm H$ and $\rm I$ in $\ce{HI}$ will be much lower than those on $\rm H$ and $\rm F$ in $\ce{HF}$. It just turns out that the increase in bond length can't compensate for this effect, hence the lower dipole moment.
$\dagger$ For those who might want to point out possible vibrational excitation or some other subtle (or not-so-subtle) effect: don't. Let's keep it simple.
answered Sep 19, 2017 at 19:45
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$\begingroup$ Point taken for no vibrations, but what about going from e.g. $\ce{CH3-NH2}$ to $\ce{CH2=NH}$? $\endgroup$– JanCommented Sep 20, 2017 at 3:17
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2$\begingroup$ @Jan Just like I said, these are different molecules with different everything. $\endgroup$ Commented Sep 20, 2017 at 4:38
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If the same point charges are separated by a larger distance, that's a higher dipole moment.
In the real world, you have to regard the distance and the actual charge difference and distribution.
In the case of HF and HI, fluorine has the highest electronegativity, which means the bond in HF is more polarised than in HI. So, in spite of the lower bond lenght, HF can have, and indeed has, a higher dipole moment.
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$\begingroup$ but why distance isnt taking into consideration ? $\endgroup$– ado sarCommented Jan 10, 2019 at 21:09
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$\begingroup$ @adosar please explain what you mean by "consideration". I don't get it. $\endgroup$– KarlCommented Jan 10, 2019 at 21:15
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$\begingroup$ Why hf has greater dipole moment if hcl bond length is bigger ? How can we know that q1d1>q2d2 only from the electronegativity difference? $\endgroup$– ado sarCommented Jan 10, 2019 at 23:47
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$\begingroup$ @adosar I explain exactly that above. It's a fact, we know the distance and the dipole moment from different measurements, so we can calculate the charge difference! And we know about the high electronegativity of fluorine, so the result makes sense. That's nature, I can't change it. $\endgroup$– KarlCommented Jan 11, 2019 at 6:14