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I was asked the following question:

How many optical isomers are there for $\ce{[Cr(NH3)2Cl2(en)]^+}$ ?

First of all I figured out the following four geometrical isomers of the above complex:

geometrical isomers of Cr(NH3)2Cl2(en)+

Out of which, (iii) and (iv) are optically inactive, due to a plane of symmetry (shown by faint line) passing through them as shown below:

visualisation of the plane of symmetry

For (iii), you can suppose: green as $\ce{en}$, orange as $\ce{NH3}$ and red as $\ce{Cl}$. Similarly, for (iv), you can suppose: green as $\ce{en}$, orange as $\ce{Cl}$ and red as $\ce{NH3}$.
Therefore only (i) and (ii) will show optical isomerism, which I've confirmed myself.


But, the following four isomers was provided as answer, in which I don't think that the second configuration of complex will show any optical activity due to plane of symmetry (as discussed above for the (iv) geometrical isomer).

enter image description here

What I am doing wrong?

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    $\begingroup$ I don't think it's you who's doing wrong. $\endgroup$ – mykhal Nov 28 '18 at 15:21
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    $\begingroup$ Well, it's a valid mirror plane. It's just that it's the same compound after mirroring... $\endgroup$ – Zhe Nov 28 '18 at 16:17
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I came to the same conclusion as you, total 4 stereoisomers (2 of them create one enantiomeric pair).

In the 3rd party solution, they depict the same isomer at bottom, i.e. only 3. The missing one is that with two $\ce{NH3}$ ligands in the same plane with $en$, and $\ce{Cl-}$ ligands in straight line (trans), perpendicular to that plane (not that horizontal, which is depicted)

Fig.0

Here is my simplified schematic representation of the octahedral complex with ethylenediamine ligand bound by back-pointing bonds in horizontal plane, hidden; the rest is vertical axis and two front-pointing horizontal bonds (seesaw geometry, not to be confused with Fischer projection for tetrahedrals!) Dotted line is mirror plane, equal sign means identity (by 180° rotation around observer-to-model-line axis).

Fig.1 Fig.2

  1. has B ligands in the (en)-defined plane, A ligands perpendicular (axis)
    (it's their bottom left isomer, idenitical with bottom right, for A=$\ce{NH3}$, B=$\ce{Cl}$; note that they have en oriented vertically, everywhere, with one end in dotted horizontal plane, other one in upper vertical position, I have it somewhat simpler, rotated for en to be horizontal);
  2. has the opposite, A's and B's swapped (and it's not mirror image to 1.)
    (it's their missing isomer);

If A and B are in the plane then there are two possibilities how to relatively place A and B in the perpendicular axis (resulting in mirror images),

  1. (it's their top right);
  2. (it's their top left).
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  • $\begingroup$ If the claimed 3rd party correct solution image you provided is complete - Yes, the solution is complete. $\endgroup$ – rv7 Nov 29 '18 at 15:05
  • $\begingroup$ @rv7 OK, I did not want to visit unknown sites. Well, you have apparently posted someone's complete solution, that is unfortunately incomplete :) $\endgroup$ – mykhal Nov 29 '18 at 15:16

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