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According to Wikipedia's article about orbital hybridisation:

[...] today it is considered an effective heuristic for rationalising the structures of organic compounds. [...] Hybrid orbitals are assumed to be mixtures of atomic orbitals, superimposed on each other in various proportions.

So are $\mathrm{sp}$, $\mathrm{sp^2}$, $\mathrm{sp^3}$ hybrid orbitals 'real'? Are they solutions of Schrodinger equation? Are they a superposition of atomic orbitals?

Mathew Mahindaratne
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Sparkler
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    The classic solutions to the orbitals for hydrogen atoms (s,p,d,... Legendre polynomials) form a complete basis set. Any wave function can be written using linear combinations of them. As for 'are they real', various STM and other surface imaging techniques show them to be amazingly accurate in some circumstances. – Jon Custer May 01 '15 at 14:35
  • @JonCuster, I was actually asking about the hybrid orbitals sp, sp2 and sp3. In Modern valence band theory it says that the molecular orbitals are LCAO and not superposition of only s and p. – Sparkler May 02 '15 at 15:32
  • From the wikipedia link it says , LCAO is a quantum superposition, which again will lead you to a dead end . Reality and Quantum mechanics just dont go well together, yet. – Gowtham May 02 '15 at 15:40
  • As far as I remember, hybrid orbitals are not real. I don't exatly remember were I read it. Maybe in Atkins or in Class Notes. – sedflix May 02 '15 at 19:58
  • Obligatory reminder that in general "orbitals" are not real - they are the solutions to mathematical models of the behavior of electrons. The electrons are real. Thus, the hybridized orbitals are no less real than molecular orbitals or atomic orbitals. They are just the solutions to a different set of equations modelling electron behavior. The better question would be Are hybridized orbitals a useful model? – Ben Norris May 03 '15 at 10:52
  • The hybrids $sp, sp^2, sp^3$, are not solutions of the Schrödinger equation in a polyelectronic atom. The atomic orbitals $s$ and $p$ are solutions, but their linear combinations are not, because their energy level are different. Not very much different I know. But if the energy levels of $s$ and of $p$ were equal, their linear combination would be solution of the Schrödinger equation. And this may only happen in the Hydrogen atom. – Maurice Feb 26 '21 at 16:13

2 Answers2

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As soon as you move to systems with more than one electron them all orbitals are approximations (i.e. not real). Hybridized orbitals are no more "heuristic" than atomic orbitals.

The transformation between hybrid and atomic orbitals can be mathematically, though not uniquely, defined (e.g. as the minimization of inter-orbital electrostatic repulsion) and the transformation does not affect the total electron density.

See for example http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html and http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html

Jan Jensen
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  • so minimization of LCAO energy is not an asymptotic approximation of actual ("real") orbitals? – Sparkler May 04 '15 at 03:39
  • No, because there are no actual ("real") orbitals – Jan Jensen May 04 '15 at 06:44
  • I'm not sure what you mean. It sounds like you're rejecting a core assumption: hydrogen AOs obtained from solving Schrodinger equation represent actual orbitals of the electron of hydrogen. – Sparkler May 05 '15 at 15:11
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    No, for hydrogen what you are calling AOs are actually wavefunctions, from which you can compute the densities which is are physical observables. It's when you start using these 1-electron functions to describe multi-electron wavefunctions that they cease to be "real". – Jan Jensen May 06 '15 at 07:39
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My understanding is the following: You can only get exact analytical solution for the Schrödinger equation for a) a single electron and b) a single proton - nucleus. That is, for the hydrogen atom with a single electron. The solution for this atom exactly describes the wavefuntion for the electron in any of the orbitals (the base one, or a excited one).

When you have more than an electron, and/or more than a nucleus, then you cannot get exact analytical solutions for the Schrödinger equation.

(I am not sure, if you have a multiproton nucleus with a single electron, e.g. He+, if you can get exact solutions; I believe so).

As for atoms with multiple electrons, you can consider that the efect of other electrons shields the charge in the nucleus, and then you can approximate the orbitals in this case to the single electron case.

When you have multiple atoms forming a molecule, and there are shared electrons positioned around two or more of their nuclei, you cannot get an exact analytical solution - even if you still had a single electron, as in H2+. But it was found that by adding or substracting the atomic orbitals, you get a reasonable approximation for the molecular orbitals that form covalent bonds. Eg. By combining the s orbitals of two H atoms, you get a close approximate solution of the molecular orbital of the H2 molecule. (Note that in these H2+ or H2 examples, we are not "using" hybridization yet.)

When you consider covalent molecules formed by atoms with electrons in different types of orbitals, e.g. s and p, then again, it was found that a reasonable approximation for the solution of the molecular orbitals was obtained by a) combining the wavefunction of the different orbitals in each atom (the ones with similar energy, symmetry, etc.) to calculate a "hybrid orbital" and b) combining the hybrid orbitals of each atom to get an approximation of the wavefunction of the molecular orbits.

Alfonso Perea Moraleda
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