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Is it possible that if the two lone pairs in a molecule occupy different orbitals, then could this make the molecule chiral? What I mean is, suppose a carbon atom is attached to two different atoms and has two lone pairs in different orbitals, then could the carbon atom be considered chiral?

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    $\begingroup$ You could say that in methane each C-H bond is formed from a different orbital, but that doesn't make methane chiral, does it? $\endgroup$ – orthocresol Oct 13 '18 at 12:54
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Chirality refers only to the positions of the nuclei. So, if we have the situation you described where one Carbon is bonded to two other atoms (e.g. CX$_2$ or CXY) it can never be chiral. (i.e. a three atom molecule can always be superimposed on its mirror image by a single rotation)

Now, why don't we consider the atomic and hybrid orbitals populations when assigning chirality? This comes down to the fact that the atomic and hybrid orbitals are just a helpful tool for rationalizing/predicting the structure and reactivity, but they are not representations of the actual wavefunction. The true wavefunction is defined by a set of ''molecular orbitals.'' These orbitals always have the same symmetry group as the nuclear coordinates. In this case, there is a reflection through the plane defined by the three atoms (some jargon: this is called the C$_{s}$ symmetry point group). So if the nuclear positions are non-chiral, so is the wavefunction.

(More on chirality and symmetry: http://symmetry.otterbein.edu/tutorial/applications.html)

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  • $\begingroup$ Thanku , but I never said the carbon has to be bonded to identical atoms , I said different ones $\endgroup$ – Chloritone_360 Oct 13 '18 at 19:15
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    $\begingroup$ Doesn’t change the situation at all. Any three atom molecule will have at least Cs symmetry and thus cannot not be chiral. Symmetry classifications are at the heart of this. $\endgroup$ – PJ R Oct 14 '18 at 4:07
  • $\begingroup$ If we have triatomic $\ce{X-Y-Z}$ and Y has 2 lone pairs and, for the sake of argument, one lone pair is in an $\ce{sp^3}$ orbital and the other lone pair is in an $\ce{sp^4}$ orbital, then isn't $\ce{X-Y-Z}$ chiral since it has no symmetry element and is non-superimposable upon its mirror image? $\endgroup$ – ron Oct 15 '18 at 19:06
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    $\begingroup$ Ron, a three atom molecule always has at least Cs symmetry because you can always draw a mirror plane (the plane that the three define). Remember that the electronic wavefunctuon (what actually is going on) always conforms to the symmetry of the nuclear coordinates. This is a case where the hybrid orbital concept doesn’t really match well with reality. $\endgroup$ – PJ R Oct 15 '18 at 19:16
  • $\begingroup$ In water the two lone pairs are in orbitals with different hybridizations (roughly p and sp), so in this case symmetry planes remain. But if we had a case (my earlier comment) where the two orbitals had different hybridizations such that the symmetry is destroyed, the, it seems to me, the molecule is no longer superimposable upon its mirror image. $\endgroup$ – ron Oct 15 '18 at 19:56
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There are quite a few molecules with two lone pairs residing in different orbitals that still are achiral. A simple example would be methanethiol, $\ce{H3CSH}$. On sulphur, due to its position in the 3rd period the best description is usually to consider the atom unhybridised, which is reflected by the bond angle being close to $90°$. This puts one lone pair in the 3s and one lone pair in the 3p orbital of sulphur.

While these two lone pair orbitals are clearly different that alone would not let the molecule be chiral: an s orbital is spherical in nature and a p orbital can be approximated by the rotation of a figure of 8. The important bit is that both feature a plane of symmetry that traverses them and further contains the carbon atom and two hydrogens. Since the molecule features a plane of symmetry, it is achiral. Yes, the p orbital is actually antisymmetric with respect to this plane of symmetry but the electron phase is not exactly an absolutely measurable property.

This thought experiment could be generalised. But I will now approach the answer from the other direction and talk about how orbitals are actually derived. They are all solutions to the Schrödinger equation $\hat H\Psi = E\Psi$. Because protons and neutrons are 2000 times heavier than electrons, we apply the Born-Oppenheimer approximation in calculation that says we can consider the position of the nuclei constant, calculate an appropriate wave function, then vary the nuclei positions slightly, see if we find a lower energy, rinse and repeat until we arrive at a minimum.

Calculating the energy requires to determine the electric potential that the nuclei are causing and see how the electrons would form standing waves in said potential.

If we temporarily ignore the methyl hydrogens (we were only interested in the lone pairs on sulphur), we only have three atoms which will always form a plane. This means that the electric potential will always have a plane of symmetry. This in turn means that the electronic structure will also feature that same plane of symmetry. But a plane of symmetry means achirality; therefore two different orbitals will still obey the underlying symmetry and will not be able to cause chirality.

There’s a third argument: we cannot distinguish individual electrons. Each electron pair is identical to each other one. Thus, two electron pairs being in different orbitals will always function like two hydrogen atoms—no chirality observed.

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  • $\begingroup$ Jan, relative to your third argument: Don't the two lone pairs have different energies, basicities and occupy different spatial locations?? $\endgroup$ – ron Oct 16 '18 at 17:04
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I think you are mixing some concepts. Completely agree with PJR: chirality is related to the positions of the nuclei in the molecule. If your argument was correct, water H2O will be also chiral.

You also should consider that "chirality" is not restricted to organic compounds, you can also find it in inorganic coordination complexes.

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    $\begingroup$ Independent of the lone pair stereochemistry, the two oxygen substituents are identical (hydrogen) in water. Hence a plane of symmetry is maintained perpendicular to the $\ce{H-O-H}$ plane and bisecting the oxygen. Consequently water could not be chiral. $\endgroup$ – ron Oct 15 '18 at 23:19
  • $\begingroup$ Of course your are right, I was thinking of HOD. coming to thing about it, the two lone pair are idnetical $\endgroup$ – PAEP Oct 17 '18 at 9:59

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