This question was in my book. According to me CO should be polar as it should have a dipole moment.

But I found that the $\sigma$-electron drift from C to O is almost nullified by the $\pi$-electron drift from O to C. What is about the $\pi$-electron drift towards the Carbon atom?

In Google it is given CO is polar, but my book says it is practically nonpolar ... now I am confused.

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    $\begingroup$ CO is polar only. it is proved by its activity to ir and microwave rays, $\endgroup$ – user19485 Aug 28 '15 at 9:14
  • $\begingroup$ @user19485 Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. At the moment this read more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer. For now I have converted your answer to a comment. $\endgroup$ – Martin - マーチン Aug 28 '15 at 10:45
  • $\begingroup$ Since the difference in the polarity caused by sigma drift and pi drift is quite small, it is practically non polar. But, chemically, it is polar in nature, as you see in the answers below. $\endgroup$ – Aneek Sep 16 '15 at 15:20

Well, it's just like you said: $\sigma$ electrons drift from C to O, because O is more electronegative, and $\pi$ electrons drift in the opposite direction, because... Look at it this way: C and O are both sp, each with a lone pair facing away from the other atom. They form a $\sigma$ bond and a $\pi$ bond, too. Now they want to form another $\pi$ bond, but C got no more electrons to use, so O has to throw in two of his own. That's the drift we are talking about.

As it happens, the two effects almost negate each other. The resulting dipole moment is so insignificant, I don't even remember which way it is.

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    $\begingroup$ This is not true. All orbitals except for the HOMO, including the two pi orbitals, are polarised towards the oxygen. The lone pair of carbon has a larger dipole moment than all of the other orbitals combined due to its size and location. $\endgroup$ – Martin - マーチン Aug 28 '15 at 10:49
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    $\begingroup$ I never said they aren't. Surely all orbitals are polarized towards O. Now, on one of these orbitals sit two electrons that were his in the first place. They belonged to O, and now they are somewhere between C and O - admittedly, closer to O, but still in between. That's the drift I'm talking about. As for the lone pair of carbon, isn't it matched by that of oxygen? $\endgroup$ – Ivan Neretin Aug 28 '15 at 11:02
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    $\begingroup$ It's not like I'm inventing a bicycle here. Look: en.wikipedia.org/wiki/Carbon_monoxide#Bonding_and_dipole_moment $\endgroup$ – Ivan Neretin Aug 28 '15 at 11:05

Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please see the question about the direction of the dipole moment.

The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment. \begin{align} \ce{{}^{\ominus}\!:C#O:^{\oplus}} && \text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} && \text{Direction:}~\longleftarrow\hspace{-1ex}+ \end{align}

Unfortunately the explanation of the bonding in carbon monoxide is far from simple, and it is perfectly understandable that you are confused. The dipole moment can be very well explained with molecular orbital theory, which I have already written about.

I will include a brief explanation at this point as I saw that the explanation on Wikipedia is from my point of view incomplete (not to say wrong).

Let's first state the obvious: Oxygen is more electro-negative than carbon. Therefore one can expect that all bonding orbitals are polarised towards the oxygen. That is of course only true when we are staying in the Lewis bonding picture. Based on this one would expect that the lone pairs cancel each other and since all orbitals of the triple bond are polarised to the oxygen, the dipole moment should also point in that direction.

This is obviously wrong, since the experiment tells us something different. There are ten valence electron in this molecule. They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements. The low lying orbitals are all polarised towards oxygen, because of electronegativity. The HOMO on the other hand is strongly polarised towards the carbon and is very large. In addition the nuclei contribute a dipole moment because of their asymmetry of charge. With less positive charge on the carbon nucleus than on the oxygen nucleus, the dipole component of the nuclear charge contribution also has its negative end on carbon. The combination of HOMO and nuclear dipoles in the reverse direction is larger than the combination of all the other occupied orbitals.

It is important to understand that it is often not possible to know about the properties of a compound simply by looking at a structure on paper. Very often the simple models we love and use break down easily for small molecules. A bonding situation has to be explained as is and not in terms of what would have been when we look at individual atoms and how does it change.

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  • $\begingroup$ ‘They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements.’ I don’t know for certain, but in the picture you showed in the other post I see MO number 4 which looks antibonding to me … $\endgroup$ – Jan Aug 28 '15 at 15:51
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    $\begingroup$ @Jan It is not, the nodal plane 'curves' through the oxygen. You can only see that when you would cut through the molecule. The same applies for the home, but it's the carbon in this case. $\endgroup$ – Martin - マーチン Aug 28 '15 at 16:24
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    $\begingroup$ @Aneek when it comes to carbon monoxide, there are no simple answers. The accepted one is wrong, but very easy to comprehend. This one is harder to understand, but correct. You can choose. $\endgroup$ – Martin - マーチン Sep 12 '15 at 7:59
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    $\begingroup$ There is actually another dipole contribution at work ... from the nuclei. Carbon has six protons there, oxygen has eight, thus the charge distribution from the nuclei is the same as a pair of +7 charges plus a dipole of +/-1 charges separated by the internuclear distance. This latter dipole, like the net dipole from the lone pairs, is negative towards carbon. $\endgroup$ – Oscar Lanzi Jun 22 at 14:46
  • $\begingroup$ @martin check my edit. We need to add the nuclear contribution because it is bigger than the net result. $\endgroup$ – Oscar Lanzi Jun 22 at 15:59

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