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Why does $\ce{C=O}$ have a larger dipole moment than $\ce{C-O}$?

According to me, dipole moment directly depends upon bond length and electronegativity difference. In $\ce{C=O}$ and $\ce{C-O}$, (I guess, please clear if I am wrong) the EN difference is same, and bond length of $\ce{C=O}$ is less than $\ce{C-O}$. So, $\ce{C=O}$ must have less dipole moment than $\ce{C-O}$ but it is not so. Why?

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    $\begingroup$ Intuitively, you can also see that in the $\ce{C=O}$ bond, you "share" two electrons between $C$ and $O$, and they both contribute, while in a $\ce{C-O}$ bon, you only share one. The increase of distance is not enough to compensate. Of course the correct interpretation is more akin to resonance, so be careful with the electron sharing images $\endgroup$
    – Martigan
    Jun 8, 2018 at 8:13
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    $\begingroup$ Always remember that EN is a pretty heuristic concept. It is good for getting a rough idea how some element will behave chemically, but one should never expect a phenomenon to depend only on bond lengths and EN values. $\endgroup$
    – leftaroundabout
    Jun 8, 2018 at 9:34

1 Answer 1

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According to Wikipedia, bond dipole moment depends on:

  1. Distance between atoms and
  2. Overall charge difference, not just electronegativity difference.

Resonance tells us that there is some amount of charge separation in $\ce{C=O}$ bonds because of the $\ce{C+-O-}$ contributor. This difference in charge, in addition to the electronegativity difference, is more significant than the decrease in distance between atoms, hence the larger dipole moment for $\ce{C=O}$.

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