Does $\ce{Ni((en)2(Cl)2)}$ show cis-trans form ?

And if No

Why not ?

  • $\begingroup$ Sounds like a homework question. Can you post what you think are the cis- and trans- forms of this compound? $\endgroup$
    – Eric Brown
    Jun 1, 2013 at 11:54
  • $\begingroup$ The question is asked whether the compound will show geometric isomerism . I think it should show cis-trans but the answer says no. I can't figure out why , are bidentate ligands always bonded to adjacent sites of the geometry ? $\endgroup$
    – user1735
    Jun 1, 2013 at 12:23
  • $\begingroup$ If you draw en, then you can see that there is no way for it to reach all the way around the molecule to bond to a trans position. (I think it's the smallest bidentate ligand.) $\endgroup$
    – Eric Brown
    Jun 1, 2013 at 12:28
  • $\begingroup$ I can draw a cis (w/r/t the Cl ligands) form, so if it does not exist (i.e. has never been observed), then why would that be (which is what the question is really asking)? $\endgroup$
    – Ben Norris
    Jun 1, 2013 at 14:54
  • $\begingroup$ Theoretically you can draw both situation; when monodentate ligands replaced with bidentate ligands cis- trans- isomers can be drawed (in ML2B2 type). How did you have an answer that says "no" ? $\endgroup$ Jun 1, 2013 at 20:35

1 Answer 1


$\ce{en}$ is bidenate so it has to take two neighbour corners from octahedron (let say equatorial position). The second $\ce{en}$ can bind with 1st nitrogen also on equator but then the 2nd has a choice: it can bind in equator or on axial position. So in theory it would be possible to see two isomers. The reality has however more to do with energy. $\ce{en}$ is very good ligand and $\ce{Ni++}$ makes very stable quadratic planar complexes. For that reason both $\ce{en}$ goes into equatorial position and the weaker ligands $\ce{Cl-}$ occupy axial positions.


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