Isomers of $\ce{[Co(en)2Cl2]^+}$

Sorry for the bad drawing and having $\ce{NH}$ instead of $\ce{NH2}$

In the first structure, the chlorine atoms are in trans position on the plane.

In the second structure, the chlorine atoms are in cis position on the plane.

In the third structure, one chlorine is in the plane and the other chlorine is occupying one axial position.

Now all three of these will have one enantiomer making the number of isomers in total $6$.

So $\ce{[Co(en)2Cl2]+}$ should have $6$ optical isomers, but the answer in the book says that $\ce{[Co(en)2Cl2]+}$ has $3$ optical isomers. How?

  • $\begingroup$ Compound 1 has a plane of symmetry and does not have enantiomers. Compounds 2 and 3 are enantiomers of each other. $\endgroup$ – Zhe Aug 22 '17 at 18:52
  • $\begingroup$ @Zhe In 2 both chlorine is on the plane and in 3, one chlorine is at axial position. So how 2 and 3 are enantiomers of each other ? I thought bonds don't break in optical isomerism. $\endgroup$ – user8277998 Aug 22 '17 at 18:56
  • 3
    $\begingroup$ (1) in octahedral geometry there is no inherent difference between "equatorial" and "axial" sites. Simply by 90° rotation of the molecule you can interconvert "equatorial" and "axial" sites. So your argument for why the second and third are not enantiomers doesn't work. (2) why don't you try mirroring the second structure and seeing whether it is the same as the third? You may have been taught to do that with organic compounds. $\endgroup$ – orthocresol Aug 22 '17 at 19:08
  • $\begingroup$ I suggest building a model... $\endgroup$ – Zhe Aug 22 '17 at 19:10
  • 3
    $\begingroup$ (3) the first structure is not an optical isomer of the other two. So it is incorrect to say that there are three optical isomers of [Co(en)2Cl2]+. (4) the first structure is not chiral and hence it does not have an enantiomer. (5) All in all, it seems you need to take it slowly and go back to basics (check every mirror image to see whether it is superimposable) instead of coming up with shortcuts of your own for determining chirality, since these shortcuts aren't giving you the correct answer. $\endgroup$ – orthocresol Aug 22 '17 at 19:15

In simplistic terms, we can see $\ce{[Co(en)2Cl2]^+}$ as an $\ce{[MA4B2]}$ pseudooctahedral complex. We can initially consider the $\ce{A\bond{->}M}$ and $\ce{B\bond{->}M}$ bonds identical. Trying this out on a model octahedron or molecular modelling kit will show you that only two possible different configurations exist: cis and trans.

Here, however, we have one of the ligand types actually being a briding ligand. Because the bridge is too short to allow for a trans configuration, the two coordination sites of these bridging ligands will always be cis to each other.

For the complex in which the two chlorido ligands are trans to each other, only one possible arrangement of the $\ce{en}$ ligands is possible. At first approximation, the entire complex should have $C_\mathrm{2v}$ symmetry if not more: a $C_2$ axis with two perpendicular $\sigma_\mathrm{v}$ planes; one bisecting the $\ce{en}$ ligands through the $\ce{C-C}$ bond, the other being orthogonal and mapping each $\ce{en}$ onto the other. Due to the presence of planes of symmetry, this complex cannot be chiral.

In the complex in which the two chlorido ligands are cis to each other, one could think of up to three potential arrangements for the bridges; however, as I pointed out, it is not possible to have the two coordinating atoms of one $\ce{en}$ ligand be trans to each other, reducing the total down to two. These two possibilities do not differ in whether one of the chlorides is in an ‘axial’ position (remember, we are assuming equal bond lengths initially so there is no defined ‘axial’ position). Rather, in one case you have a leftwards-turning screw and in the other case a rightwards-turning screw if you assume the $\ce{en}$ ligand to be a plane and wanted to screw the entire thing into your paper.

Neither of these two structures have a plane of symmetry (or higher symmetry) and both are therefore optically active. Careful analysis will reveal, that they are in fact mirror images and thus enantiomers of one another. They are called the Δ and Λ isomers, respectively. All three isomers are attached in the image below.

the three possible isomers of dichloridobisethylendiamincobalt(III)
Figure 1: isomers of $\ce{[Co(en)2Cl2]^+}$. From left to right: trans, cis-Λ, cis-Δ.

Thus, there are three isomers of the compound, two of which are optical isomers of each other (the third is a geometrical isomer).

Finally on the topic of whether axial bonds are actually longer, i.e. whether there is a Jahn-Teller type distortion or not. The answer is no. $\ce{Co^{III}}$ is a $\mathrm{d^6}$ complex. It could be assumed to be high or low spin. In case it accidentally ends up to be low spin, the $\mathrm{t_{2g}}$ orbitals are fully populated while the $\mathrm{e_g}$ orbitals are entirely unpopulated; there is no need for Jahn-Teller distortion.

In the potentially more likely high-spin case, one of the $\mathrm{t_{2g}}$ orbitals is fully populated while all other orbitals are half-populated. We might expect a Jahn-Teller distortion to remove the degeneracy of the $\mathrm{t_{2g}}$ orbitals. However, this distortion would be minor at best: the $\mathrm{t_{2g}}$ orbitals do not interact well with the ligands since they are oriented to point in-between the ligand orbitals. (On a group theory basis, the $\mathrm{t_{2g}}$ orbitals are nonbonding.) Therefore, any symmetry reducing effect will be very minor at best and not lead to a noticeable distortion. Thus, it is reasonable to assume near-identical bond lengths and no principal (‘axial’) direction.

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