enter image description here

The molecule I've drawn has neither a plane of symmetry nor a center of symmetry, even though some books refer to this as optically inactive.

Can you please elaborate why is this so, or those books are wrong.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. We prefer to not use MathJax in the title field, see here for details. I actually have no idea what your question is, have you already read the article on Wikipedia? $\endgroup$ May 24, 2017 at 6:47
  • $\begingroup$ @Martin - マーチンthanks for information, probably I referred some wrong artical $\endgroup$ May 24, 2017 at 7:02

2 Answers 2


Ethylenediammine or $en$ is an symmetrical didentate ligand. You can remember a rule of thumb for this if you like:

In the case of $3$ symmetrical didentate ligands, i.e. $[\ce{M(AA)3}]$, the compound exists in only one form, which is optically active

So the complex you've mentioned will be optically active, no doubt.

Also, as a reference, I'm quoting the wikipedia article at https://en.wikipedia.org/wiki/Tris(ethylenediamine)cobalt(III)_chloride#Stereochemistry

The complex can be resolved into enantiomers that are described as $\Delta$ and $\Lambda$.

The only symmetry elements that can be found in $\ce{[Co(en)3]^3+}$ are the three $C_2$ axes of rotation, one of which is shown in this image (colored in red) on wikipedia: https://upload.wikimedia.org/wikipedia/commons/e/e5/Coen3lel3.png

enter image description here

As a side note, here's another rule of thumb:

In the case of $3$ unsymmetrical didentate ligands, i.e. $[\ce{M(AB)3}]$, the compound exists in two forms, namely fac (stands for facial) and mer (stands for meridional), and both of these forms are optically active.

Important note: There's a minute error in this answer; and that's the distinction between optically activity and chirality: I used the word "optically active" while it should've been "chiral". It's bad I missed it, but I'd like you to read Ben Norris' comment on this below. I'm leaving the answer untouched, editing it rip the comment of its value. I encourage you to learn from my mistake.

  • 6
    $\begingroup$ This answer (and the other one) correctly explain why $\ce{[Co(en)3]^3+}$ is chiral. However, there is a fine distinction between "chiral" and "optically active". A substance or solution is optically active if it rotates plane-polarized light. The solution needs chiral molecules to do this, but not all solutions of chiral compounds are automatically optically active. A racemic mixture is not since it contains equal amounts of both enantiomers. $\ce{[Co(en)3]^3+}$ could be considered optically inactive if either the enantiomers cannot be separated or the enantiomers interconvert in solution. $\endgroup$
    – Ben Norris
    May 24, 2017 at 11:17
  • $\begingroup$ @Ben Norris naturally so :) $\endgroup$
    – porphyrin
    May 24, 2017 at 13:14

Some molecules can be chiral even if they do not have an asymmetric carbon. In these molecules, such as the one you draw, it is the overall shape of the molecule that is important. There has to be no centre of inversion or mirror planes (including rotation reflection $S_n$). In point group terms these mirror planes are called vertical, horizontal, dihedral and a mirror plane is part of the rotation reflection operation. They are labelled $\sigma_v,\, \sigma_h,\, \sigma_d,\, \text{and } S_n$ respectively). Your molecule follows this rule in that it has a 3-fold axis and three 2-fold axes at right angles to the 3-fold axes. In terms of point groups this is called $D_3$. There are two enantiometric forms.

Chiral molecules belong to the point groups $C_n, \, \text{and } D_n ,\;n=1,2,3\cdots$.

The figure shows the symmetry elements. You can see that here is no centre of inversion or mirror planes. Look at www.molecule-viewer.com for examples of point groups and molecules of all different point groups. An explanation of how molecules rotate plane polarised light is summarised in the answer to this question; Dependence of the angle of rotation on the wavelength of plane polarized light

d3 molecule


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