Question #c4cb7

Let's use 1,2-dichloroethene as our example.

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Note that if both chlorines were attached to the same carbon, there would a net dipole. There would be no "cis" or "trans" for the molecule. Also, note that there is a greater net dipole for #C-Cl# than there is for #C-H#.

Let's say we cut the molecule down the middle of the double bond to have top and bottom halves. If you notice with trans, we have both Cl and H on both the top and bottom. If we were to put the dipoles into a math equation, we can say "#(C-H)+(C-Cl)=(C-H)+(C-Cl)#", which would sum up as #0=0#. In other words, there is no net dipole because the chlorines are cis to each other.

Let's look at cis now. using the math equation from before, we would have "#(C-H)+(C-H)=(C-Cl)+(C-Cl)#", which would be considered a false equation. There would be a net dipole that would be going down towards the chlorines.

So, geometry of the molecule around the double bond determines whether there is a net dipole or not. You can only have two of two substituents for there to be no dipole (e.g. 2 -H & 2 -Cl, or 2 -F & 2 -OH). Additionally, the same substituent has to be diagonal across the double bond (i.e. trans-chloro or trans-fluoro) to cancel any net dipole.