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Why is carbon dioxide non-polar every explanation keeps using the symmetry argument but I want to know what is fundamentally cancelling out because as far as I can tell there should be a positive middle without two negatives on the outside?

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    $\begingroup$ Exact duplicate from Chemistry StackExchange : Why is carbon dioxide nonpolar? $\endgroup$ – user14250 Feb 18 '20 at 8:36
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    $\begingroup$ Because it's quadrupolar! $\endgroup$ – Anger Density Feb 18 '20 at 13:35
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    $\begingroup$ The symmetry argument IS the argument. You should fiddle around with symmetries and geometry untill you see that.... You cannot have a dipole when you have a centre of inversion. It is as simple as that. No vehicles that have less than 3 wheels are cars, no goats that are female have horns, it is just observable facts. Having a CoI means you cannot have a dipole. $\endgroup$ – Stian Yttervik Feb 19 '20 at 14:40
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Yes it is about the partial charges on the atoms but the dipole is a vector not just the charge distribution which is positive in the center and negative on the O's but the vectors cancel

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    $\begingroup$ Using the O=C=O notations as if it's a geometrical representation is misleading. H-O-H is polar. The fundamental reason is sp3 hybridization. The 4 bonds of carbon are all identical. In H-O-H, oxygen has 2 pairs of unshared electrons and two bonds where oxygen contributes one of the electrons. $\endgroup$ – MSalters Feb 18 '20 at 11:31
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    $\begingroup$ yes but this is the 2D geometrical representation of CO2 for H2O it is bent $\endgroup$ – ChemEng Feb 18 '20 at 19:28
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    $\begingroup$ This is the best answer. However, it could be improved by adding a similar picture of a polar molecule also with three atoms but a net dipole (e.g. H2O). $\endgroup$ – DrSheldon Feb 18 '20 at 22:17
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    $\begingroup$ @MSalters It really should be looked at the other way. The symmetry of the molecule is the fundamental reason. Hybridization is just a model to try to explain and predict symmetry of molecules. $\endgroup$ – Stian Yttervik Feb 19 '20 at 14:33
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The difficulty might be due to nomenclature : How is polarity defined?

In physics the electric field of any distribution of charges can be decomposed into a series of electric multipole moments. The most fundamental is the monopole which means that the resultant electric charge is non-zero. Chemists use the term ionic. If the distribution is overall neutral but there is a separation between the centres of positive and negative charge then there is a dipole.

Having a dipole moment is sometimes taken as being synonymous with being polar - see first answer to Are asymmetric molecules necessarily polar? By this definition $CO_2$ is a non-polar molecule because its overall dipole moment is zero.

However, close to one side of the $CO_2$ molecule the nearer of the two dipoles will dominate, so the molecule will have a non-zero effect on an external charge. The 2 local dipoles (2x2) constitute a linear electric quadrupole. See also Difference between Non-Polar and Dipole moment $\vec\mu$=0. The $H_2O$ molecule has both a dipole and a non-linear quadrupole.

Higher moments are possible. 4 dipoles (4x2) pointing from the centre to the corners of a square form a octupole. The multipole components are generally weaker by an order of magnitude in succession. While the electric field of the monopole falls off as $1/r^2$ that of the dipole, quadrupole and octupole fall off as $1/r^3, 1/r^4, 1/r^5$ respectively.

Chemists do not use the multipole expansion; instead they define a polarity index.

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  • $\begingroup$ The polarity index is not used by chemist discussing the electric dipole moment of a molecule, to reassure that things are the same. It is only on the specific context of interactions that confusion may arise. $\endgroup$ – Alchimista Feb 18 '20 at 9:01
  • $\begingroup$ Does *pole expansion matter here? *pole is about EM field far away from the distribution, but for molecure we care mainly about near field which decides structures and reactions. $\endgroup$ – jw_ Feb 20 '20 at 4:12
  • $\begingroup$ @jw_ That is a good point. The multipole expansion might not be useful for calculating fields around molecules. But it does explain why charge distributions which have no dipole moment can still be polar, and it predict that higher order moments are progressively weaker and therefore signigicant only at ever closer range. $\endgroup$ – sammy gerbil Feb 20 '20 at 5:05
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Linear B-A-B molecules cannot be polar. i.e. cannot have a permanent electric dipole for symmetry reasons. Stated plainly, the electronic displacement from A atom towards the two B atoms (or the other way around if A is more electronegative than B atoms) must be symmetric with respect to the central atom, ending up with a molecule without a net electric dipole.

The only way an AB$_2$ molecule can have a permanent electric dipole, is if its equilibrium position is bent. This is not the case of CO$_2$, but there are other molecules,like SrCl$_2$ o SO$_2$ which do have a bent equilibrium configuration and consequently an electric dipole.

The basic reason why some of these molecules are linear and others are bent can be traced back to the interplay between a coulombic effect between ions, favoring linear geometry, and the possibility that a high electronic polarizability may favor a bent configuration.

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    $\begingroup$ Your explanation of bent vs. linear molecular geometry is inadequate. $\endgroup$ – my2cts Feb 18 '20 at 0:30
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    $\begingroup$ Actually, the geometry purely depends on the hybridization of the central atom and the number of electron lone pairs present. $\endgroup$ – Sam Feb 18 '20 at 4:10
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    $\begingroup$ @Sam hybridization and lone pairs are just a couple of QM ingredients which concur to the ionic core polarizability. They are not alternative explanations. QM level tells the electrons what to do. But, at the end of the day, all chemical bonds depend on the fundamental electrostatic interaction. I am stressing the electrostatic level. BTW, it should be the most adapted to SE Physics. $\endgroup$ – GiorgioP Feb 18 '20 at 7:23
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    $\begingroup$ @my2cts It's your right to disagree. However, without explanation, it is neither a scientific comment nor a helpful contribution to this community. I have worked and published a few papers on this subject and I think I know what I am speaking about. I cannot exclude that I may be wrong, but I would like to know why. Your feeling without arguments is not enough. $\endgroup$ – GiorgioP Feb 18 '20 at 14:18
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    $\begingroup$ This sounds like a circular argument. Straight molecules are straight because they're not polar, and they're not polar because they're straight. $\endgroup$ – MSalters Feb 18 '20 at 16:24
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In $\text {CO}_2$ the central carbon atom is $sp$ hybridised and hence has a linear shape. Now the $ \text O$ atoms on both side have higher electronegativity and hence the electron density is partially shifted towards $\text O$ atoms. This leads to two separate dipoles of equal magnitude and opposite direction in parts of the atom but the molecule as a whole has $0$ net dipole. Molecules with zero dipole moment are non-polar and so is $\text {CO}_2$. Had it been that $\text {CO}_2$ had $sp^2$ hybridised carbon atom then it would be a polar one (Note that $\text {CO}_2$ cannot have $sp^2$ hybridised carbon atom).

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In the chemistry dialect polar means having a non zero electric dipole moment. That is, if you put your molecule in a uniform electric field, it will orient itself aligning its dipole moment to it.

Let's consider an arbitrary molecular structure with no symmetry: it is likely that it will show some dipole moment. But if we stick that to its mirror image in such a way that the dipole moments have the same direction but opposite verses, the result will be a molecule that, globally, does not care about an external electric field (unless this becomes strong enough to mess with its internal structure).

I know that the symmetry argument is not very satisfying the first times it is brought to the table, but once you grasp the idea, it became so powerful and ubiquitous that it is always the first thing you look for in a new problem!

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  • $\begingroup$ The symmetric argument is pretty good if you point out that the dipole acts like a vector. Only the null vector is equal to itself when rotated and/or mirrored. $\endgroup$ – MSalters Feb 18 '20 at 16:29
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Basic covalent/chemical bonding, the two oxygen atoms are completing the valence shell of the carbon atom, this is stable due to the repulsive forces of the opposite charges being weaker.

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  • $\begingroup$ My follow up question is will the molecules of carbon dioxide feel a force when dissolved in water from the dipoles of water since from what I understand stand from the replies charges will still be present a either side of the molecule. $\endgroup$ – Luke Feb 20 '20 at 18:42

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