# Is a tetrahedral complex with two symmetric bidentate ligands optically active?

If the shape of of a complex $$\ce{M(AA)2}$$ (with $$\ce{AA}$$, a bidentate ligand) is tetrahedral, then is it optically active?

Currently I know that it is optically inactive, but I want to know the reason for this. About which plane is this symmetrical?

• The plane which passes through two A and between the other two A, for example. Oct 3, 2021 at 8:17
• @Ivan Neretin please elaborate your reasoning Oct 3, 2021 at 10:55
• To OP: you should use $\ce{...}$ specifically for chemical formulae. Plain dollar signs $...$ are to be used for mathematical expressions. What's the difference? Well, for one, the former is upright and the latter italics. Upright $\ce{I}$ $\ce{I}$ is the symbol for iodine, but italics $I$ $I$ represents something else entirely, like inertia, nuclear spin, or the identity matrix. Please read: FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? for more information. Oct 3, 2021 at 12:01

Get in contact with other (more senior) students/TA's/your instructor and get access to a molecule kit (an example, no affiliation) to build a model of these complexes. If you can't, use small cubes cut of a potatoes (as atoms) and toothpicks (as bonds). Then build these complexes.

An idealized representation of a tetrahedral complex around a central ion with two bidentate ligands is shown in the animation below: carbon atoms in black, nitrogen in blue, the transition metal in ocher:

As mentioned by @Ivan, it is possible to define a plane across two atoms of nitrogen and the central ion to bisect the molecule in hemispheres like a mirror. A second one is just $$90^\circ{}$$ to the one shown, too. Whenever you identify at least one of such a mirror plane across the whole molecule, the complex/the molecule/the object is achiral. And if a molecule is achiral, it is not optical active. In your study of chemistry, you are going to encounter this later on again with meso compounds in organic chemistry, e.g., tartaric acid.

For the sake of clarity of the representation, I omitted to display the hydrogen atoms. Second, independent of the former, was to a assume the five membered rings as flat. While such a conformation is not much likely,
it may be -- depending on what you aim to highlight in molecular symmetry -- a permissible simplification.

(A nice approach to train oneself recognizing symmetry in molecules is, beside using molecule kits, the set of pages around Symmetry@Otterbein.)

• Your animation was was very helpful... Oct 4, 2021 at 12:45
• +1 for the movie! Oct 4, 2021 at 14:12