organic chemistry - Difference of dipole moments of dichloromethane and trichloromethane?

Why is the dipole moment of $\ce{CH2Cl2}$ ($1.60 ~\mathrm{D}$) greater than that of $\ce{CHCl3}$ ($1.08~\mathrm{D}$)?

Based on my knowledge of vectors, I feel it should be the other way around.

Answer

Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.

Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$ where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.

For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$ which will be $1A$, lesser than in the case of dichloromethane.

To visualise how can you get those angles and cosines, see this page.