# Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

I know that dipole moment depends upon the strength of dipole and angle between the dipoles.

1,1-Dichloroethylene has two chlorine atoms with 120° angle between them because of $$\mathrm{sp^2}$$ hybridization of carbon atom.

cis-1,2-Dichloroethylene has the same with the angle between chlorine atoms as 60° (or else the dipole moment of trans form would not have been zero).

So, the dipole moment of cis should be greater than that of 1,1-dichloroethylene, but the answer says the opposite (μ(1,1-dichloroethylene) > μ(cis-1,2-dichloroethylene)). Can anyone help me?

• The 1,1 compound has the larger dipole. The dipole is charge times distance, so as well as calculating the component of the charge along each axis you also have to consider the distance apart of the charges. The total is the sum of each part. May 8, 2020 at 15:01

Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:

For convenience to identify, I put $$a$$, $$b$$, $$c$$, and $$d$$ as the magnitudes of applying dipole moments of relevant $$\ce{C-Cl}$$ bonds. Yet they are theoretically identical: $$a = b = c = d = \mu$$.

Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $$a$$ and $$b$$, $$a \sin 30^\circ$$ and $$-b \sin 30^\circ$$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $$a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$$. This is the net dipole moment of cis-1,2-dichloroethene.

Now, consider 1,1-dichloroethene. The vertical components of vectors $$c$$ and $$d$$, $$-c \cos 30^\circ$$ and $$d \cos 30^\circ$$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $$c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$$. This is the net dipole moment of 1,1-dichloroethene.

Therefore it is clearly, $$\sqrt{3}\mu \gt \mu$$. Realistically, $$\mu_{1,1}$$ is smaller than $$\mu_{1,2}$$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $$\mu_{1,1} = \mu = \pu{1.3 D}$$ and hence, $$\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$$, which is actually $$\pu{1.9 D}$$.

• +1 I like that you actually showed the vectors broken down into perpendicular components and did the math.
– MaxW
May 9, 2020 at 16:49
• Yes, you are correct. And I even checked the values for confirmation, but the book states the reason that u11>u12 to be resonance. They state that in u11 due to resonating structures, there will be charge involved in dipole moment, whereas not that effective in u12 Sep 15, 2020 at 15:26

cis-1,2-Dichloroethene has the larger dipole.

cis-1,2-Dichloroethene, 1.9 D

The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 1.9 D.

1,1-dichloroethylene, 1.3 D

The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.

trans-1,2-Dichloroethene, 0 D

The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 0 D.