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Why does $\ce{CH2Cl2}$ have a higher dipole moment than $\ce{CHCl3}$?

My argument being that the $\ce{CCl3-}$ ion being rather stable, can generate a good amount of polarity in the bond, moreover, shouldnt the dipoles of both $\ce{Cl}$ atoms cancel out?

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  • $\begingroup$ The dipoles of both Cl atoms aren't pointing in the exactly opposite directions, hence they cancel out only partially. $\endgroup$ Mar 23, 2017 at 7:56
  • $\begingroup$ In that case, wouldnt the addition of one more such dipole add to the net resultant vector @Ivan Neretin $\endgroup$
    – SubZero
    Mar 23, 2017 at 9:14
  • $\begingroup$ The said dipole is not quite aligned with the sum of the first two, so the resulting change could be either way. As it happens, it is actually in favor of CH2Cl2. If you are familiar with vectors and 3D geometry, you can check it yourself. $\endgroup$ Mar 23, 2017 at 9:21
  • $\begingroup$ Don't use mhchem in question titles. $\endgroup$ Mar 23, 2017 at 11:10

2 Answers 2

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As the dipolar moment is a vectorial property, you have to take into account both the number of chlorine atoms around the carbon atom, as well as their relative arrangement -- among them, as well in respect to the carbon atom.

Lacking access to experimentally determined data about the molecular geometries,you may sketch the molecules in question, for example with Avogadro, and optimize the geometry (here, MMFF94 was deployed). You are literally able "to walk around" the molecule and its dipolar moment.

enter image description here

You easily will recognize a feature in common -- the similar direction of the resulting dipolar moment. Simultaneously, you will recognize the different angle defined by Cl-C-Cl, too. However, beside the angle, distances like C-Cl and C-H are affected by the different steric demand and electronic properties, too; equally influencing the overall dipole moment. Hence the molecule of dichloromethane is more polar than the one of chloroform.

Sidenote: a force field like MMFF94 is a computationally cheap approach to optimize molecular structures. There are better methods, but more costly; for the level of the question addressed, however, I thought it already provides the insight necessary here.

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In $\ce{CHCl3}$ the dipole moment of the $\ce{C-Cl}$ bond is towards $\ce{Cl}$. Since it has a tetrahedral geometry and the dipole moment is a vector quantity, the vector sum of all dipole moments would try to cancel out. As they are in the outward direction, they will cancel to some extent. While in $\ce{CH2Cl2}$, the $\ce{C-H}$ bond has the dipole towards $\ce{C}$ and there are only two $\ce{C-Cl}$ bonds. So they wouldn't cancel out.

And talking about the $\ce{CCl3-}$ ion. The carbanion would form only in certain (extreme) conditions.

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  • $\begingroup$ Sorry. My mistake $\endgroup$
    – Ava
    Mar 23, 2017 at 8:22
  • $\begingroup$ How exactly would 3 dipole vectors oriented like a tetrahedron (3 vectors with a common origin give a tetrahedron) cancel out each other $\endgroup$
    – SubZero
    Mar 23, 2017 at 9:13
  • $\begingroup$ PS. 4 vectors with common origin give tetrahedron. And I didn't say they completely cancel out.they try to $\endgroup$
    – Ava
    Mar 23, 2017 at 10:31
  • $\begingroup$ Im sorry, but we need just 3, and point accepted, that they only try to, then why dont they result in a dipole moment larget than what would arise due to 2 vectors trying to cancel each other $\endgroup$
    – SubZero
    Mar 23, 2017 at 10:37
  • $\begingroup$ I can't attach the image, but search on Google tetrahedral geometry. You would see that it requires 4. Also the angle between any two bonds is approx 109° so if you would try to do vector sum of three vectors mutually aligned at 109° and two at 109°, the first case would give lesser resultant. And the increase in H also contributes to more dipole $\endgroup$
    – Ava
    Mar 24, 2017 at 11:10

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