Draw the four isomers of C2H2Cl2Br2

I have to draw the 4 four isomers for the molecule $\ce {C2H2Cl2}$ when $\ce {Br2}$ is added to it. At last the four isomers for $\ce {C2H2Cl2Br2}$.
I have found three of them: 


![enter image description here](https://i.stack.imgur.com/8VZ06.jpg) 


I can't find the last one.




You have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)). 


This isomer is known as "meso compound". 


Meso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.


![enter image description here](https://i.stack.imgur.com/NUcIz.gif)


 4

Like Lighthart already mentioned in the comments, it is important to know which compound was used in the addition reaction. Because of the [reaction mechanism](http://en.wikipedia.org/wiki/Halogen_addition_reaction#Reaction_mechanism), only certain products are formed, this is very specific. LDC3 also explained that in the comments, but let me dissect this a little further.



> 
> The first structure is from geminal,
> 
> 
> 


In the scheme below, this is reaction $(1)$ and it forms an achiral product.



> 
> the second is from cis 
> 
> 
> 


This is depicted in $(2)$ and it forms the meso compound described in Yomen Atassis's answer.



> 
> and the third is trans (or do I have cis and trans backwards). 
> 
> 
> 


*(No you have not.)* This is depicted in $(3)$, but in this case a racemic mixture of the (*S,S*)- and (*R,R*)-product is formed, depending on which carbon atom in the bromonium ion the nucleophilic attack takes place.



> 
> The only other isomer I can think of is 1,1-dibromo-2,2-dichloroethane, but it is not made from adding $\ce{Br2}$ to dichloroethene. 
> 
> 
> 


For completeness, this is depicted in $3$.


I could further think of another isomer, which is like the above not formed via addition, it is depicted in $(5)$. Note that these last two reactions also yield achiral products.


![enter image description here](https://i.stack.imgur.com/mSLdW.jpg)


 1