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In a $\ce{CO_2}$ molecule, a total of four electron pairs are shared between the carbon and oxygen atoms, such that 2 pairs are shared between the carbon atom and each oxygen atom. Oxygen has a greater electronegativity hence the atoms of oxygen will spend greater time with shared pairs and are partially negative. Since both oxygen atoms will exhibit this same behavior, the carbon atom will become partially positive.

I was reading about the dipole moment of $\ce{CO2}$ (e.g., as here), in which it said that the net polarity of the $\ce{CO_2}$ molecule is zero because if we add the individual dipole moment vectors directed from the positive pole (carbon) to the negative poles (the oxygen atoms), the resultant will be zero which implies the polarity of the molecule is zero.

My problem is this:

If these individual dipole moments (vectors) are showing the opposite and equal pull of shared pairs of electrons toward more electronegative atoms (oxygen atoms), then the carbon atom should stay partially positive with the oxygens remaining partially negative.

Given this, though, it seems that the molecule should remain polar on whole. This is because these dipole moment vectors are acting on different shared pairs: the first vector acts on the two shared electron pairs between the carbon atom and one of the oxygen atoms, and the other acts on the two shared electron pairs between the carbon atom and the other oxygen atom. According to the following law, then, these vectors cannot cancel each other and thus $\vec u \neq\vec 0$:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

My best attempt to resolve this contradiction:

Since in reality it is the case that the net polarity of $\ce{CO_2}$ is null, I think that both vectors (viz., the 'pull' from both oxygen atoms) must be acting individually on all electrons of carbon atom and therefore they cancel each other out resulting in molecule to be non-polar on whole. This explanation isn't all that satisfying to me, however.

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    $\begingroup$ Also, see here. $\endgroup$ – hBy2Py Mar 24 '16 at 16:52
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Your confusion appears to arise from trying to consider dipole vectors as representing forces, which is incorrect.

Dipole vectors don't "act on" anything; they describe the displacement of the "center-of-charge" of a given region of space (including both the electron cloud and the nuclei) from a particular reference point. In particular, I believe you are confusing dipole moment vectors with electric field vectors:

  • The electric field vector at a point describes the force experienced there by a point charge
  • The dipole moment vector at a point describes the displacement of the "center of charge" of the system from that point

Thus, whereas the electric field contributes to the forces present in a system, the dipole is simply a post facto descriptor of a system.


Further, in this case the law you quoted is not applicable:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

It only makes sense to calculate an overall dipole moment if you use the same reference point for all of the 'subsystem' dipole vectors. Thus, in this example of $\ce{CO2}$, the two $\ce{C-O}$ 'subsystem' vectors point in exact opposite directions from the same reference point, since $\ce{CO2}$ is linear. Thus, since the outside atoms are both oxygens, the magnitudes of both 'subsystem' vectors are equal and their sum is the zero vector.


For a non-electroneutral region of space, by center of charge I mean the charge-weighted centroid of the system: (1) Choose a reference point $\vec r_0$. (2) At each point $\vec r$ in the region of interest, multiply the charge at that point, $\rho\!\left(\vec r\right)$, by the distance from the reference point, $\vec r - \vec r_0$. (3) Integrate the resulting product, $\rho\!\left(\vec r\right)\cdot\left(\vec r - \vec r_0\right)$, over all space. (4) Divide by the net charge of the region of interest. Dipole moment calculations are more complicated for electroneutral systems, but the principle is similar.

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  • $\begingroup$ In that case, carbon atom should become more positive in $\rm{CO_2}$ as compared to the carbon atom in $\rm{CO}$ molecule. And I think the dipole moments are forces is not all wrong because they can be thought as the forces exerted by somewhat negative pole(s) to shift the electron density towards themselves. This is the idea I got from Khan academy and one more lecture on youtube. $\endgroup$ – Sufyan Naeem Mar 24 '16 at 19:46
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    $\begingroup$ @SufyanNaeem Dipole moments are not forces. Period. They reflect the outcome of forces, yes, but they are not themselves forces. They are effect, not cause. $\endgroup$ – hBy2Py Mar 24 '16 at 19:53
  • $\begingroup$ Well I think I am closer but please tell me more on what is displacement of center-of-charge or suggest me a book that could tell me about that. I thin, I have tou understand this thing only. $\endgroup$ – Sufyan Naeem Mar 24 '16 at 19:57
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    $\begingroup$ @SufyanNaeem Please see my further edits $\endgroup$ – hBy2Py Mar 24 '16 at 20:20
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More than once I have complained about how ill-defined the concept of polarity is. One of the most common ways to attest whether a molecule is polar or not is to judge it form its dipole moment. When you look at the equilibrium geometry of carbon dioxide, which has $D_\mathrm{\infty h}$ symmetry, you will see, that it is impossible to have an overall dipole moment due to the inversion centre, i.e. the carbon centre.
Due to the simple picture it paints, we often decompose the total dipole moment into bond dipole moments, because this gives us a handle on how polarised these bonds are. Often enough we can estimate an overall dipole for a molecule from bond dipole moments of similar molecules. However, the overall dipole moment is just one way to express how the electron density and the field of the nuclei differ from each other, and the decomposition into different vectors is arbitrary, a mathematical model if you wish.
You can look at the dipole moment $\mathbf{q}$ as the expectation value of the dipole operator $\mathbf{P}$ from the total wave function $\Phi_0$. $$\langle\mathbf{q}\rangle = \langle\Phi_o|\mathbf{P}|\Phi_0\rangle$$

But let's have a closer look at carbon dioxide and what you have written.

In a $\ce{CO2}$ molecule, a total of four electron pairs are shared between the carbon and oxygen atoms, such that 2 pairs are shared between the carbon atom and each oxygen atom.

This is not true. This is the misconception you have when looking at a Lewis type bonding picture. If you employ the molecular orbital theory, you can see that there are actually twelve electrons in bonding orbitals, and four electrons in anti-bonding orbitals. Not counting core electrons. All of those electrons are delocalised over the whole molecule.

pi orbitals of CO2
sigma bonding orbitals of CO2
(Taken from an earlier post of mine. There is an error, MO 8 should be MO 7. BP86/cc-pVTZ level of theory. Click images to enlarge.)

If these individual dipole moments (vectors) are showing the opposite and equal pull of shared pairs of electrons toward more electronegative atoms (oxygen atoms), then the carbon atom should stay partially positive with the oxygens remaining partially negative.

This is true. This charge separation is also the reason for the quadrupole moment. Taken from BP86/cc-pVTZ:

 Traceless Quadrupole moment (field-independent basis, Debye-Ang):
   XX=              1.3594   YY=              1.3594   ZZ=             -2.7187
   XY=              0.0000   XZ=              0.0000   YZ=              0.0000

Probably a more known approach is to look at the partial charges on the same level of theory using different analysis tools.

      Mulliken     Natural        Hirshfeld     Atoms in
                   Population                   Molecules
1  C    0.336381     0.91266        0.295332      2.111095
2  O   -0.168190    -0.45633       -0.147682     -1.055548
3  O   -0.168190    -0.45633       -0.147682     -1.055547

You see, you are quite right. There is charge separation. Which mainly stems from the lone pair electrons at the oxygen atoms, i.e MO 6 and 7 (8 in the picture).
This is why I am very hesitant of calling a molecule like carbon dioxide unpolar.

Given this, though, it seems that the molecule should remain polar on whole. This is because these dipole moment vectors are acting on different shared pairs: the first vector acts on the two shared electron pairs between the carbon atom and one of the oxygen atoms, and the other acts on the two shared electron pairs between the carbon atom and the other oxygen atom. According to the following law, then, these vectors cannot cancel each other and thus $\vec u \neq\vec 0$:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

Electrons are indistinguishable, so are electron pairs. You cannot treat the electrons of a molecule as different bodies. You can only treat the molecule as a whole. The positive charged nuclei act on all electrons at the same time, all electrons are correlated. In the Born-Oppenheimer approximation you have a cloud of electrons action upon a fixed field of attractors (nuclei).

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The dipole moment results from the center of "gravity" of all negative charges and the center of gravity of all positive charges being spatially separated. Its actually the product of the positive times the negative net charges times the distance vector between their centres.

Thus a dipole moment of zero actually means that the centres of gravity coincide (unless there are no charges at all), and it does not mean that there are no net (partial) charges around. So in deed in CO$_2$ the center of gravity of all positive charges is at the location of Carbon nucleus, but for reasons of symmetry so is the center of gravity of all the negative charges (electrons).

In contrast to a zero dipole moment CO$_2$ has a substantial electric quadrupole moment, this is maybe what your intuition suggests.

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