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In my book, the order of dipole moment for CH3Cl, CH2Cl2, CHCl3 and CCl4 is given as:

CH3Cl>CH2Cl2>CHCl3>CCl4

The way I understand this is—

In case of CCl4, tetrahedral symmetry exists. Therefore, the dipole moment of three C-Cl bonds on one side, gives a resultant moment that is equal and opposite to the dipole moment of the single C-Cl bond on the opposite side. So, $\ce{\mu=0}$.

This explains CHCl3>CCl4.

Again, in case of CHCl3, the three C-Cl bonds point to one side, so a moment equal to C-Cl bond is generated as a result. A moment of C-H bond is added to it. In CH2Cl2, the moment of two C-Cl bond is added at $\ce{109^\circ}$(approx.) so the resultant is greater than on C-Cl bond.

This explains CH2Cl2>CCl4

In case of CH3Cl, the dipole moment of one C-Cl bond is added with moment of one C-H bond (resultant of three C-H bonds).

But this should mean moment of CHCl3$\ce{\approx}$ moment of CH3Cl, but clearly this is not so. Where did I go wrong?

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marked as duplicate by Mithoron, Todd Minehardt, airhuff, a-cyclohexane-molecule, pentavalentcarbon Jan 15 '18 at 1:32

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  • $\begingroup$ "But this should mean moment of CHCl3≈≈ moment of CH3Cl" In making this assumption, did you take into account that the direction of the dipole moment in a $\ce{C-H}$ bond is opposite to that in $\ce{C-Cl}$ bond? $\endgroup$ – Gaurang Tandon Jan 14 '18 at 15:33
  • $\begingroup$ @Gaurang Direction of dipole moment for C-H bond is in the same direction as the resultant due to 3 C-Cl bonds. C attracts electron from H, and Cl attracts electrons form C. So, the moments are added. $\endgroup$ – Shoubhik Raj Maiti Jan 14 '18 at 16:48
  • $\begingroup$ chemistry.stackexchange.com/questions/48066/… $\endgroup$ – Mithoron Jan 14 '18 at 22:57