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In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

However, aren't there infinitely many regions of space in which the total probability of finding the electron is equal? And if so, wouldn't that imply that it is not possible to assign a shape to an atomic orbit as there are infinitely many shapes with equal likelihood to contain the electron?

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  • $\begingroup$ I think you are confusing atomic orbitals with the graphical representation of atomic orbitals. Also I wouldn't worry too much about orbitals. It could be the single most overused concept in chemistry thanks to a poor understanding by most people that argue with them. But if you are really interested, you can read up on the solution of the Schrödinger equation for hydrogen-like atoms and Hartree-Fock theory, but this is complicated stuff if you're still in school $\endgroup$ – AMT Jan 11 '17 at 15:25
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In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

This is not correct.

An atomic orbital is a solution to Schrödinger's equation given the electrostatic potential. In the simplest case, we solve the one electron atom and perturb that with more electrons. An exact solution to the multielectron problem is intractable due to the chaotic nature of 3-body systems.

The concept you are referring to is a boundary surface. (See this question). Basically, it's a surface that has equal probability density for every point on the surface, and the surface encloses a region of space where the sum of probabilities is the desired value.

Because the surface is defined to have equal probability everywhere, and the wave function is smooth, it should be unique.

The boundary surface is usually how we represent the atomic orbitals that are familiar to us ($s, p, d, f$) but also molecular orbitals. Usually, we take something at 90% probability, for example, and plot the boundary, giving the orbital "shape." Otherwise, visualization can be tricky because the orbital only goes to zero probability at infinite distance from the nucleus, so there is a nonzero chance that an electron will be quite far from the nucleus.

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    $\begingroup$ I see, so the crucial condition I was missing is that every point on the boundary surface has the same probability density. Thank you :) $\endgroup$ – MinecraftShamrock Jan 9 '17 at 22:37
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Electrons are in quantum mechanics described by wavefunctions $\Psi(\vec{r})$. These are complex physical objects which contain all the information about the state of the electrons. The square of the wave-function $|\Psi(\vec{r})|^2$ gives the probability of finding the electron at position $\vec{r}$. As you said orbitals are often considered as the smallest volume in which the electrons are expected to be with a 90% probability. To construct these one constructs isosurfaces of the mentioned probability density and tries to choose the isosurface with the smallest volume which however still includes 90% of the electrons, i.e. the integral over the probability density over the volume of the isosurface is 90%.

Of course you could construct as many possible shapes as you want, you don't have to rely on isosurfaces. However isosurfaces provide much more physical insight, since they also include information about the gradient of the probability density function which is always perpendicular to the isosurfaces. So from just looking at isosurface you get a good insight into how the function actually looks like. This information is included in atomic orbitals. While when you would choose random shapes it would give you pretty much zero information...

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  • $\begingroup$ Thank you for this interesting and more mathematical explanation :) $\endgroup$ – MinecraftShamrock Jan 9 '17 at 23:19
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According to a stone-age paper, Chem. Rev., 1986, 86 (4), pp 681–696,

Atomic orbitals are widely referred to by chemists, but are actually only loosely defined. Hydrogenic or- bitals are too different from the optimum orbitals for describing an atom to furnish a suitable definition. Slater originally introduced STO’s as approximations to atomic orbitals when the parameters n and {where chosen by certain empirical rules. Consequently, STOs are still frequently confused with atomic orbitals. In this paper we will use the term “atomic orbital” to refer to a Hartree-Fock orbital resulting from a converged (i.e., basis set limit Roothaan-Hartree-Fock or nu- merical SCF) calculation on some state of the atom with spin (singlet, doublet ,... ), symmetry (S, P ,... ) and equivalence restrictions (e.g., px = py = pz). Unfortu- nately, this definition is still somewhat state dependent as the 2p Hartree-Fock orbital for the carbon s2p2 3P ground state is not identical to the 2p Hartree-Fock orbital for the s1p3 configuration average.

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    $\begingroup$ That information was already well seasoned then. The work of Slater, Hartree and Fock is mostly from the 20ies to 50ies, none of them was even alive any more in 1986! $\endgroup$ – Karl Jan 10 '17 at 0:27

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