10
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

Since you know about vectors, then the piece you are missing is probably the proper three-dimensional structure of the molecules.

Trichloromethane (chloroform)

Structure of chloroform

Notice that the third chlorine atom in trichloromethane is pointing away from the other two. More precisely, the three chlorine atoms are at the bottom three points of a tetrahedron. Thus the $x$ and $y$ components of the individual $\ce{C-Cl}$ bond dipoles cancel, leaving only the $z$ component.

Dichloromethane

Structure of DCM

For dichloromethane, the $y$ components fully cancel, but the $x$ components do not.

$\endgroup$
  • $\begingroup$ But how do you know that the $x$ components in Dicholoromethane sum up to be greater than the $z$ component in Chloroform? $\endgroup$ – Apurv Mar 7 '14 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.