How does B2Cl4 have two perpendicular c2' axes? I have been looking at the symmetry of $\ce{B2Cl4}$ and was cannot understand how it has two perpendicular $C_2'$ axes? 

I understand it has a $C_2(z)$, $S_4(z)$ and two dihedral planes bisecting each of the $\ce{Cl-B-Cl}$ bonds. However my book says it also has an additional two $C_2'$ axes in the $x$ and $y$ planes centred at the middle of the $\ce{B-B}$ bond. I just can't see it. Any help? ( I understand it will be difficult online but I would very much appreciate your assistance.
Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups.

In this case, a picture tells more than I thousand words I think.



The two $\ce{C_2^{'}}$ axis are turned by 45° in comparison to the C-H bonds. It therefore also has two dihedral planes, thus belonging to the $\ce{D_{nd}}$ point group.

If you have difficulties picturing this in your head that is not strange. Many people make mistakes in assigning point groups. You get a better visualization by practicing online.