While passing a plane of symmetry through any compound inorganic or organic should it contain a minimum number of atoms or can it be passed without containing any atoms ?


Yes , the plane of symmetry need not cross/pass through any atoms. The compound that you've drawn can show cis-trans isomerism. I believe you're talking about the cis form.

  1. The cis form will have a plane of symmetry
  2. The trans form will not have a plane of symmetry
  • $\begingroup$ While the principle information (plane of symmetry does not have to pass through atoms) is correct, it is wrong that cis-1,2-dichlorocyclohexan will have such a plane. The ring is in a chair conformation with one chlorine atom equatorial, the other axial (in the ground state). See also this related question: Is cis-1,2-dimethylcyclohexane a meso compound? $\endgroup$ Oct 9 '18 at 9:23
  • $\begingroup$ @Martin-マーチン While what you say is correct, it is also incomplete. The interconversion rate of the conformers might matter. In other words, the molecule might have an effective plane of symmetry on average if the conformers interconvert fast enough. $\endgroup$
    – matt_black
    Oct 9 '18 at 10:32
  • $\begingroup$ @matt_black I seriously doubt that, as far as I remember, there is no symmetrical transition state that would interconvert the two. Also, symmetry is not something you can measure, or is it? $\endgroup$ Oct 9 '18 at 10:42
  • $\begingroup$ @Martin-マーチン I don't know whether this compound interconverts or is locked in a single configuration at normal conditions. But, if it interconverts, what I said is correct. The key question is whether the chlorines are distinguishable under normal conditions. $\endgroup$
    – matt_black
    Oct 9 '18 at 10:49
  • $\begingroup$ @matt_black As with the linked question, the molecule will most definitely interconvert, even at low temperatures. I have trouble understanding what average symmetry is, or how it could have an effective plane of symmetry. Even if it interconverts ultrafast, you would be able to observe both enantiomers simultaneously. If you mean with effective plane of symmetry that the molecule appears achiral, then that is true, but wouldn't then every racemic mixture have this property? Anyway, this is not the place to discuss that (maybe Chemistry Chat), I just pointed out a quite common mistake in the answer. $\endgroup$ Oct 9 '18 at 11:29

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