Non polar molecule with polar bonds?

[neilparker62]

TL;DR Summary

Is it possible for a non-polar molecule to have polar bonds ?

Consider for example Carbon Dioxide. Oxygen is more electronegative than carbon so should obtain the "lion's share" of the paired electrons in the double bonds. But (as I see it anyway) the oxygen atoms on either side of the central carbon "assist" the carbon atom to maintain an even share of the paired electrons. So the molecule as a whole is non polar covalent.

But now can we still say the individual c=o bonds are polar ? According to tables of electronegativity they are but is this a "moot point" given that in-situ the electrons are in fact evenly shared ? I'm asking this in the context of multiple choice questions which ask for molecules containing non-polar covalent bonds. Methane is one of the choices but is apparently a wrong choice. Is that really so or is there an incorrect understanding of C-H bonds in the specific context of methane.

 

[TeethWhitener]

The polarity of a molecule (whether it’s dipole moment is nonzero) is more a function of molecular geometry than polarity of individual bonds. So carbon tetrafluoride is nonpolar in virtue of its tetrahedral symmetry, even though the individual CF bonds are highly polar.

Generally the CH bond is about as close to nonpolar as a heteroatomic bond gets. I suppose there’s a slight polarity, but not appreciable for most organic reactivity considerations. A truly nonpolar bond is only homoatomic, and only between symmetry equivalent atoms. So the CC bond in ethanol is polar, even though it is between two carbons. The CC bond in ethylene glycol, however, is nonpolar on average (at least over a time period where intermolecular interactions and intramolecular motions can be ignored).

 

[apostolosdt]

Methane, as I assume you know, gets a tetrahedral structure after the various atomic orbitals are mixed to become hybrid molecular orbitals; while carbon dioxide becomes linear. Yes, the two carbon atoms on either side of oxygen offer some kind of "balance", so the molecule's polarity appears to be zero, but the two double bonds must oscillate a bit with respect to their polarity. That's Pauling's resonance theory, right? So, yes, each double bond, seen separately, should periodically exhibit some minute polarity.

But I guess that's not a good answer for a multiple choice paper!

 

[neilparker62]

The polarity of a molecule (whether it’s dipole moment is nonzero) is more a function of molecular geometry than polarity of individual bonds. So carbon tetrafluoride is nonpolar in virtue of its tetrahedral symmetry, even though the individual CF bonds are highly polar.

Generally the CH bond is about as close to nonpolar as a heteroatomic bond gets. I suppose there’s a slight polarity, but not appreciable for most organic reactivity considerations. A truly nonpolar bond is only homoatomic, and only between symmetry equivalent atoms. So the CC bond in ethanol is polar, even though it is between two carbons. The CC bond in ethylene glycol, however, is nonpolar on average (at least over a time period where intermolecular interactions and intramolecular motions can be ignored).

I understand that CF bonds are "individually" highly polar. But in the context of the carbon tetrafluoride molecule, the strong pull (on the shared electron pair) by any single fluorine atom is 'mitigated' by 3 other fluorine atoms symmetrically placed around the central carbon atom. So we can't really talk about the individual bonds being "highly polar" any more. They all become non polar creating non polar carbon tetrafluoride.

I take the point from @apostolosdt about the polarity oscillating but can we presume the oscillation is essentially about a zero polarity "ground line" ? And would also happen even in say the H-H bond in hydrogen molecules.

 

[TeethWhitener]

I understand that CF bonds are "individually" highly polar. But in the context of the carbon tetrafluoride molecule, the strong pull (on the shared electron pair) by any single fluorine atom is 'mitigated' by 3 other fluorine atoms symmetrically placed around the central carbon atom. So we can't really talk about the individual bonds being "highly polar" any more. They all become non polar creating non polar carbon tetrafluoride.

I take the point from @apostolosdt about the polarity oscillating but can we presume the oscillation is essentially about a zero polarity "ground line" ? And would also happen even in say the H-H bond in hydrogen molecules.

This becomes relevant in infrared spectroscopy, where the absorption is a function of the derivative of the molecular dipole moment with respect to an internal coordinate. The high polarity of, for example, the C=O double bond in CO₂ versus the relatively low polarity of the C=C bond in allene (H₂C=C=CH₂) means that the absorbance of the IR-active bands in CO₂ will be proportionally stronger than in allene. But also note that this is still dependent on the symmetry of the molecule (and the symmetry of the molecular motion being excited in that particular IR absorbance band).

 

[snorkack]

IR activity depends on the derivative of molecular dipole moment with respect to distance. Are there any molecules/bonds which are, not precisely transparent (that would be too much of an exact coincidence) but appreciably polar yet only weakly IR active... because although the dipole moment is large, its derivative with respect to bond length, near the equilibrium bond length, is small?

 

[TeethWhitener]

IR activity depends on the derivative of molecular dipole moment with respect to distance. Are there any molecules/bonds which are, not precisely transparent (that would be too much of an exact coincidence) but appreciably polar yet only weakly IR active... because although the dipole moment is large, its derivative with respect to bond length, near the equilibrium bond length, is small?

There are certainly normal modes of polar bonds in polyatomic (>2) molecules which are IR inactive because of symmetry. The symmetric stretch of CO2 is an example.

For a general molecule, the IR selection rule is that the the direct product of the irreducible representations of the initial state, final state, and dipole operator must contain the totally symmetric representation for the transition integral to be nonzero. But technically this is only a necessary condition, not a sufficient one. In math:
If
$$\langle f | \mu | i \rangle\neq0$$
Then
$$\Gamma^{(f)}\times\Gamma^{(\mu)}\times\Gamma^{(i)}=A_1$$
But it’s not an iff statement (as far as I’m aware). So I suppose what you’re suggesting could be possible, but I don’t know of any particular examples.

Edit: the selection rule is normally presented as the contrapositive of what I wrote above. That is, if the direct product of the irreps mentioned doesn’t contain the totally symmetric representation, the transition integral must be identically zero.

 

[DrClaude]

IR activity depends on the derivative of molecular dipole moment with respect to distance. Are there any molecules/bonds which are, not precisely transparent (that would be too much of an exact coincidence) but appreciably polar yet only weakly IR active... because although the dipole moment is large, its derivative with respect to bond length, near the equilibrium bond length, is small?

To add to @TeethWhitener informative answer: dipole moment is charge times distance. So the dipole moment comes about from a separation of partial charges across a bond. To get a polar bond with no dependence with bond length would require the partial charges to be also proportional to bond length, which would be strange.

 

[snorkack]

To add to @TeethWhitener informative answer: dipole moment is charge times distance. So the dipole moment comes about from a separation of partial charges across a bond. To get a polar bond with no dependence with bond length would require the partial charges to be also proportional to bond length, which would be strange.

Consider the limit cases of dipole moment.

1. Since the total charges of nuclei are finite, it follows that partial charges are also finite. Therefore at the limit of zero bond length the dipole moment must go to zero.
2. For diatomic molecules, all electron affinities, even the biggest (Cl), are smaller than all ionization energies, even the smallest (Cs). So even CsCl will, on the limit of infinite bond length, split into neutral atoms, not ions. Now what happens on the approach, at large bond lengths? Theoretically the partial charges might shrink more slowly than the distance grows so that although the charge converges to zero, the dipole moment diverges to infinity; but I suspect it would converge faster so that the dipole moment also converges to zero when partial charge does.

If the dipole moment is zero for both zero and infinite bond length, there must be some bond length for which the dipole moment goes through maximum. And at that length, the derivative of dipole moment in respect to distance must be zero.
Are there any bonds for which the equilibrium bond length around which they oscillate is coincidentally close to the maximum of dipole moment? (Exactly equal would be too much of a coincidence, as suggested before).

 

[TeethWhitener]

I agree with @DrClaude that it would be strange, and as I mentioned before, I don’t know of any examples, but it’s not strictly forbidden as far as I know. My guess is that most bonds are pretty far from their max dipole moment (in that you can probably stretch them and increase the dipole moment quite a bit). But I don’t know for sure. It’s probably something you could mess around with in a quantum chemistry program if you had the inclination.

 

[neilparker62]

Apologies for re-opening this thread after so long. I am still trying to get my head around how a non polar covalent molecule can have polar covalent bonds other than nominally so (difference in electronegativities). An example being boron trifluoride. If the boron - fluorine bonds are polar covalent , each of the 3 fluorine atoms must have a partial negative charge. Alternatively the "pull" on the shared electrons from the other 2 fluorine atoms "cancels" each partial negative charge. But you can't have your cake and eat it - the physical distribution of charge within the (nominally) polar bonds has then shifted such that they are no longer polar.

 

[TeethWhitener]

But you can't have your cake and eat it - the physical distribution of charge within the (nominally) polar bonds has then shifted such that they are no longer polar.

Why doesn't this argument work equally well against HCl?

Edit: the actual physical distribution of charge in BF3 is such that there is a higher electron density on the fluorines than on the boron. The boron has a partial positive charge of $+\delta$ and each fluorine has a partial negative charge of $-\frac{1}{3}\delta$. (There are some subtleties here about what spatial volume "belongs" to each atom, but the point is that the electron density is not distributed symmetrically across the line between the boron nucleus and a fluorine nucleus)

 

[neilparker62]

Why doesn't this argument work equally well against HCl?

Edit: the actual physical distribution of charge in BF3 is such that there is a higher electron density on the fluorines than on the boron. The boron has a partial positive charge of $+\delta$ and each fluorine has a partial negative charge of $-\frac{1}{3}\delta$. (There are some subtleties here about what spatial volume "belongs" to each atom, but the point is that the electron density is not distributed symmetrically across the line between the boron nucleus and a fluorine nucleus)

HCl is polar covalent. The H is partially positive and the Cl partially negative. I thought boron trifluoride was non polar covalent but you seem to be indicating that it is in fact polar covalent since shared electron pairs are pulled towards the fluorine atoms.

 

[TeethWhitener]

HCl is polar covalent. The H is partially positive and the Cl partially negative. I thought boron trifluoride was non polar covalent but you seem to be indicating that it is in fact polar covalent since shared electron pairs are pulled towards the fluorine atoms.

Yes, electron density is pulled toward the fluorines (thus each B-F bond is a dipole). But because of the symmetry of the molecule, the three B-F dipoles cancel out and the molecule as a whole is nonpolar.

Imagine you have two concentric spheres, one is negatively charged and the other is positively charged. The dipole moment of the entire configuration is zero, but if you look at any radius by itself, the dipole moment will be non-zero. It’s the same idea with a molecule like BF3 or CO2. Each bond is a dipole, but all the dipoles together are oriented such that there is no net dipole.

 

[TeethWhitener]

Found this image off a google search that is a good illustration.
https://homework.study.com/explanat...tructure-for-bf3-and-explain-your-answer.html
Red means higher electron density and blue means lower electron density. There is lower electron density at the boron than out toward the fluorines. Thus each B-F bond is individually a dipole. But put them together as in BF3 and they cancel, meaning that BF3 is nonpolar.

 

[neilparker62]

I think I may be struggling with several misconceptions.

Firstly I am not sure I quite understand the nature of a "dipole moment". My understanding of forces on bound electrons is that they are fundamentally electrostatic in nature. Thus coulombic as in $\frac{k q_1 q_2}{r^2}$. So the first thing is where does q r (dipole moment) come from ??

Secondly in the pic showing electron density, do the red areas represent high probability location of electrons in the shared pair or of all outer shell electrons in fluorine ?

If we work with coulombic forces, the electrons in each shared pair experience coulomb forces from the central boron , from the "sharing" fluorine (closest/strongest) and from the two other fluorine nuclei (further away / weaker). So that whilst normally one would expect the shared pair to be pulled towards the strongly electrophilic fluorine atom (creating regions of partial charge) , the 'counter pull' from the other two fluorine nuclei along with the central boron nucleus results in a situation whereby all 3 shared pairs are evenly shared. Hence no partial charges resulting in a non polar covalent molecule with non polar covalent bonds.

 

[snorkack]

I think I may be struggling with several misconceptions.

Firstly I am not sure I quite understand the nature of a "dipole moment". My understanding of forces on bound electrons is that they are fundamentally electrostatic in nature. Thus coulombic as in $\frac{k q_1 q_2}{r^2}$. So the first thing is where does q r (dipole moment) come from ??

Look at a test charge q at some distance r from a pair of charges Q₁ and Q₂, such that Q₁=-Q₂, and the distance between the charges is d, so that the distance q to Q₁ is r and the distance q to Q₂ is r+d.
Then the coulombic forces are, due to Q₁
F₁=kqQ₁/r²
due to Q₂
F₂=kqQ₂/(r+d)²=-kqQ₁/(r+d)²
Adding them
F=kqQ₁(1/r²-1/(r+d)²)
How else do you express the expression
(1/r²-1/(r+d)²)?
Bring it to common denominator...
((r+d)²-r²)/((r+d)²*r²)=(r²+2rd+d²-r²)/((r+d)²*r²)=(2rd+d²)/((r+d)²*r²)
So for large r compared to d, you are left with an expression that approaches Qd

If we work with coulombic forces, the electrons in each shared pair experience coulomb forces from the central boron , from the "sharing" fluorine (closest/strongest) and from the two other fluorine nuclei (further away / weaker). So that whilst normally one would expect the shared pair to be pulled towards the strongly electrophilic fluorine atom (creating regions of partial charge) , the 'counter pull' from the other two fluorine nuclei along with the central boron nucleus results in a situation whereby all 3 shared pairs are evenly shared. Hence no partial charges resulting in a non polar covalent molecule with non polar covalent bonds.

No, it does not. The nearby fluorine still has a stronger pull than the distant fluorines combined. Electrons still concentrate near fluorines and away from boron.

 

[TeethWhitener]

@snorkack gave you a decent answer, although I'll reiterate that your argument works equally well for HCl as it does for a BF bond. I don't know why you accept the argument in one case and not in another. Heck, just imagine "F" is actually "H" and "BF2" is actually "Cl" and they're the exact same argument.

Secondly in the pic showing electron density, do the red areas represent high probability location of electrons in the shared pair or of all outer shell electrons in fluorine ?

It's just electron density. Electrons are indistinguishable. You can't say whether one electron is electron x or electron y, only what all of them cumulatively are doing. It sounds like you're working off a Lewis pair model. That's fine for some things, but for figuring out where electrons will go based on coulombic forces, it's not really useful.

 

[neilparker62]

Look at a test charge q at some distance r from a pair of charges Q₁ and Q₂, such that Q₁=-Q₂, and the distance between the charges is d, so that the distance q to Q₁ is r and the distance q to Q₂ is r+d.
Then the coulombic forces are, due to Q₁
F₁=kqQ₁/r²
due to Q₂
F₂=kqQ₂/(r+d)²=-kqQ₁/(r+d)²
Adding them
F=kqQ₁(1/r²-1/(r+d)²)
How else do you express the expression
(1/r²-1/(r+d)²)?
Bring it to common denominator...
((r+d)²-r²)/((r+d)²*r²)=(r²+2rd+d²-r²)/((r+d)²*r²)=(2rd+d²)/((r+d)²*r²)
So for large r compared to d, you are left with an expression that approaches Qd

Many thanks - now I'm beginning to grasp how "dipole moment" relates to coulomb force or electric field at some point P distance r from the dipole.. The expression you end up with seems to have $r^3$ in the denominator - do I understand correctly that the electric field due to a dipole is $\frac{kq\vec{d}}{r^3}=\frac{k\vec{p}}{r^3}$. With d<<r. Only thing I can't quite see is what happened to the factor '2' in 2rd ?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html

 

[neilparker62]

@snorkack gave you a decent answer, although I'll reiterate that your argument works equally well for HCl as it does for a BF bond. I don't know why you accept the argument in one case and not in another. Heck, just imagine "F" is actually "H" and "BF2" is actually "Cl" and they're the exact same argument.

It's just electron density. Electrons are indistinguishable. You can't say whether one electron is electron x or electron y, only what all of them cumulatively are doing. It sounds like you're working off a Lewis pair model. That's fine for some things, but for figuring out where electrons will go based on coulombic forces, it's not really useful.

As mentioned I had no idea what "dipole moment" meant physically. I hope I do now. I had thought that non polar covalent meant no regions of partial charge. Whereas actually it means no net electrical field on account of a zero sum for dipole moments. If I understand correctly ??

 

[TeethWhitener]

As mentioned I had no idea what "dipole moment" meant physically. I hope I do now. I had thought that non polar covalent meant no regions of partial charge. Whereas actually it means no net electrical field on account of a zero sum for dipole moments. If I understand correctly ??

Two things: First, a decent E&M textbook will derive the inverse cube dependence of the far field behavior of a dipole.

Second, addressing the question above: if you stand far enough away from a molecule like BF3 and perform a multipole expansion (wiki) of the electric field, the result will not have a dipole term. This doesn't mean the electric field is zero. It just means that the lowest order term (assuming the molecule overall is uncharged) will be at best a quadrupole.

 

[neilparker62]

Two things: First, a decent E&M textbook will derive the inverse cube dependence of the far field behavior of a dipole.

Second, addressing the question above: if you stand far enough away from a molecule like BF3 and perform a multipole expansion (wiki) of the electric field, the result will not have a dipole term. This doesn't mean the electric field is zero. It just means that the lowest order term (assuming the molecule overall is uncharged) will be at best a quadrupole.

Thanks. I also found an online derivation here.

https://byjus.com/physics/dipole-electric-field/

Along axis electric field does have in it the factor of 2 I wasn't sure about from @snorkack s derivation.

One last question: is there a database of dipole moments for various molecules? As well as more 'granular' data on individual dipole moments within a molecule. Eg C-H bond in methane and other alkanes. I found references one with C d- and H d+ and another with C d+ and H d- for methane.

Thanks for all the assistance - I've certainly learnt a lot.

 

[apostolosdt]

Have you tried handbooks, like the CRC Handbook of Chemistry and Physics?

 

[TeethWhitener]

NIST Chemistry webbook (Webbook.nist.gov) and CCCBDB (https://cccbdb.nist.gov/) are also good resources.

 

[neilparker62]

Thanks for the references. I had a look at the CCCBDB database for methane and carbon tetrachloride. The dipole moment data just showed zero by every method used. Not unexpected of course, but the information I am looking for is the experimental dipole moments of individual C-H bonds in methane and individual C-Cl bonds in carbon tetrachloride. I want to understand to what extent (if at all) individual dipole moments are affected by other dipole moments in the same molecule.

 

[TeethWhitener]

Thanks for the references. I had a look at the CCCBDB database for methane and carbon tetrachloride. The dipole moment data just showed zero by every method used. Not unexpected of course, but the information I am looking for is the experimental dipole moments of individual C-H bonds in methane and individual C-Cl bonds in carbon tetrachloride. I want to understand to what extent (if at all) individual dipole moments are affected by other dipole moments in the same molecule.

This question itself is a little fraught. Even if you knew the exact electron density of the molecule, you would still have to answer the question of what part of that density belongs to which atom, which is not trivial. Different answers will give you different dipole moments. Probably the closest you’re going to get experimentally is the IR oscillator strength for a vibrational mode, but even then, the normal modes are only simple two-atom stretches in the crudest approximation. In reality they’re complicated atomic motions distributed over the entire molecule.

 

[neilparker62]

In that case, does it make any sense at all to talk about individually polar bonds in a non polar molecule ?

Thinking simplistically , I imagine a shared electron pair in (say) $CCl_4$ to be pulled towards the Chlorine atom. That's in the context of a single bond. But in the molecule as a whole that shared pair is also influenced by the other shared pairs arriving at a situation whereby the molecule as a whole is non polar.

If measurement of the polarity of single bonds is effectively impossible, under what pretext can we claim that there are "polar covalent bonds in a non polar molecule" ? Even in $CCl_4$ let alone $CH_4$. Other than by recourse to subtraction of electronegativities which may not validly represent the physics of the shared electron pairs in the molecule as a whole.

 

[apostolosdt]

OK, it's a looong time ago when I attended a class on chemistry, but I remember the textbook we were studying from, "Chemical Principles" by Dickerson et al., had a couple of chapters on how the electronic density 'lobes' were defining the properties of the molecules. I remember the excellent diagrams and figures. That was the first edition, now must be at least in the third, so I can't guarantee that those chapters survived. Otherwise, as perhaps already mentioned, there is still the classical treatise by Pauling on chemical bonds.

 

[neilparker62]

OK, it's a looong time ago when I attended a class on chemistry, but I remember the textbook we were studying from, "Chemical Principles" by Dickerson et al., had a couple of chapters on how the electronic density 'lobes' were defining the properties of the molecules. I remember the excellent diagrams and figures. That was the first edition, now must be at least in the third, so I can't guarantee that those chapters survived. Otherwise, as perhaps already mentioned, there is still the classical treatise by Pauling on chemical bonds.

Ok - thanks. Lot of reading to do!

https://authors.library.caltech.edu/25050/13/Chapter_12.pdf

 

[TeethWhitener]

In that case, does it make any sense at all to talk about individually polar bonds in a non polar molecule ?

Thinking simplistically , I imagine a shared electron pair in (say) $CCl_4$ to be pulled towards the Chlorine atom. That's in the context of a single bond. But in the molecule as a whole that shared pair is also influenced by the other shared pairs arriving at a situation whereby the molecule as a whole is non polar.

If measurement of the polarity of single bonds is effectively impossible, under what pretext can we claim that there are "polar covalent bonds in a non polar molecule" ? Even in $CCl_4$ let alone $CH_4$. Other than by recourse to subtraction of electronegativities which may not validly represent the physics of the shared electron pairs in the molecule as a whole.

We seem to be going in circles. It makes as much sense as you want it to. Using CCl4 as an example, it’s clear that electron density is nonuniform and higher near the Cl nuclei than the C nucleus. So treating each C-Cl bond by itself (and excluding the other Cl’s), there is clearly a dipole moment, although making this quantitative is a lot of work (see Bader’s atoms in molecules). It serves as a useful approximation/heuristic, especially in certain applications. To take another example: the normal modes of CH3Cl (chloromethane) that involve Cl and C can be expected to have a much higher oscillator strength than those that mainly involve only C and H, which is reflected in the IR spectrum. But again, this is an approximation, as any normal mode will likely contain contributions from all the atoms in the molecule.

 

[DrClaude]

Maybe looking up the Mulliken population analysis will help you.

 

[neilparker62]

We seem to be going in circles. It makes as much sense as you want it to. Using CCl4 as an example, it’s clear that electron density is nonuniform and higher near the Cl nuclei than the C nucleus. So treating each C-Cl bond by itself (and excluding the other Cl’s), there is clearly a dipole moment, although making this quantitative is a lot of work (see Bader’s atoms in molecules). It serves as a useful approximation/heuristic, especially in certain applications. To take another example: the normal modes of CH3Cl (chloromethane) that involve Cl and C can be expected to have a much higher oscillator strength than those that mainly involve only C and H, which is reflected in the IR spectrum. But again, this is an approximation, as any normal mode will likely contain contributions from all the atoms in the molecule.

Thanks for your patience - greatly appreciated. I think I am generally out of my depth here but scouring some (intimidating) references on Mullikan populations, I came across the following set of graphs which (if (big if!) I interpret correctly) is quite telling in respect of carbon tetrachloride. +- 0 charge on Carbon atom ?!

 

[neilparker62]

Maybe looking up the Mulliken population analysis will help you.

Thanks - way out of depth here but did some "cherry picking" - see graphs above.

 

[TeethWhitener]

This is probably as close as you’re going to get for what you want. Bader’s AIM might also be worth a look. Mulliken population analysis looks specifically at the populations of the atomic orbitals in the basis set used to build up the molecular orbitals, so again, this is a particular way of divvying up which electron density belongs to which atom (i.e., it’s up to you to decide what the nature of those atomic orbitals is and how many of them to include). As I said, it’s an approximation, since in reality all the electrons “belong” to all the atoms (atomic orbitals don’t really exist by themselves in a molecule), and different approximations will give different answers.

 

[DrDu]

If measurement of the polarity of single bonds is effectively impossible, under what pretext can we claim that there are "polar covalent bonds in a non polar molecule" ? Even in $CCl_4$ let alone $CH_4$. Other than by recourse to subtraction of electronegativities which may not validly represent the physics of the shared electron pairs in the molecule as a whole.

Today, it is possible to map the electron distribution in molecules precisely using highly resolved X-ray diffraction methods. NMR shifts also yield information about electronic density on the individual atoms. Polarity of the bonds in symmetric molecules gives rise to non-vanishing higher multipole moments, e.g. a high quadrupole moment in CO2. We are not living in the nineteenth century.