There are three different possible structures (known as isomers) of a dibromoethene molecule, C2H2Br2C2H2Br2 . One of them has no net dipole moment, but the other two do. Draw Lewis structures for each of these structures. Include HH atoms.

Consider the structure of this compound. A carbon-carbon double bond connects two carbon atoms. There are two hydrogen atoms and two bromine atoms attached to the two carbon atoms. The two bromine atoms might be on the same side of the double bond. Alternatively, they may be on opposite sides of the double bond. There are thus two structural isomers with this molecular formula:

1,1-dibromoethene, where the two bromine atoms are on the same side of the double bond; as well as
1,2-dibromoethene, where the two bromine atoms are on opposite sides of the double bond.

The C-Br bond is polar. There's no way that the dipole due to these bonds will balance each other in 1,1-dibromoethene.

However, these dipoles might indeed balance each other in 1,2-dibromoethene. The two bromine atoms might be either on the same side of the molecule (structure 2) or on opposite sides along the diagonal (structure 1). Besides, unlike groups on two ends of a single bond, groups on the two ends of a C=C double bond cannot rotate freely along the bonding axis. As a result, structure 1 and 2 cannot interconvert without breaking bonds. Structures 1 and 2 are configurational geometric isomers.

The dipoles due to the two C-Br bond balance each other along the diagonal in structure. Thus there's no net dipole in structure 1.

The dipoles due to the two C-Br bonds will not balance each other in structure 2. As a result, there will be net dipole in structure 2.