Is plane of symmetry a necessary and sufficient condition for optical inactivity in coordination compounds? [closed]

Question

In organic chemistry, we have to check for all three types of symmetry in a compound to check its chirality, ie, plane, alternate axis and centre of symmetry. But in coordination compounds, do we need to check all the symmetries to check its chirality? The ligands are supposed to be achiral. The difference between the two being that in coordination compounds only one "chiral" center is present but in organic compounds it can be more than one.

All the achiral coordination compounds of any coordination number which I have come across have a plane of symmetry. But for those which don't have, I have to check whether it's image coincides with itself or not. So, is plane of symmetry a necessary and sufficient condition for achirality of coordination compounds?

This thing is necessary for my purpose, as I have to solve problems in limited time, so I cannot use the orignal method always.

Answer 1

The number $\pi$ for bowhead whales^† is still 3.1415... Same thing here.

Chirality of coordination compounds is the same thing as chirality of organic compounds. If all your symmetry elements are just rotations, you have it; if some are improper rotations, then you don't. It is as simple as that, and is not going to be any simpler.

Can you just check for the planes and call it a day? Well, you kinda can, and that will do, until you come across a compound which doesn't have a plane, but still is achiral because of some other symmetry element.

Are there such compounds? Well, that's like asking whether there is a place on Earth where three bricks stand on top of each other. You may google long and hard, or you may apply for a travel grant to visit countries with rich traditions of brickwork masonry... or you may just take three bricks and put them on top of each other. That's it; problem solved.

Here are my bricks. I claim that the tetrahedral $\ce{[Zn(NH3)4}]^{2+}$ has an inversion axis $\bar4$ and hence is achiral. You may point out that it has a plane as well. Fine. Let's give it some substituents: change each ligand to $\ce{N(CH3)3}$. What? Does it still have a plane, given that everything which rotates is rotated to the appropriate position? Fine. Let's give it different substituents. Change each ligand to $\ce{CH3NHC2H5}$. Where is the plane now?

So it goes.

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† There was an old anecdote about an even more old paper on biophysics which was said to include some volume calculations and some crude approximations of a whale as a cylinder, then of course the number $\pi$ showed up, followed by the hilarious phrase along the lines of "The number $\pi$ for bowhead whales is about 3.14".

Answer 2

No. A plane of symmetry is a sufficient, but not necessary criterion to exclude optical activity of a molecular compound. Equally sufficient, but not necessary criterion for the absence of chirality is the presence of a centre of inversion. Note that you probe for these criteria independent from each other the whole molecule.

An easier example of organic molecules suitable for the first criterion is (2R,3S)-tartaric acid - equally known as meso-compound - where you may draw a mirror plane across the molecule even if you have two examples of atom centred chirality in the molecule present. (It is not good to call the stereogenic atoms «chiral atoms». Beside atom centred chirality, there equally are helical and axial chirality, too.)

By training, you are going to identify these compounds by inspection of their formulae:

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Don't hesitate to build models of the molecule in question and to redraw the structure formulae like the following to get more familiar with the topic:

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Only the two first columns of the four depict a mesocompound.