Can electrostatic forces explain electron configurations?

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Main Question or Discussion Point

I've been studying basic atomic structure-- shells, subshells, orbitals, the four quantum numbers, the periodic table, etc.

I've seen diagrams of "atomic structure" that show arrangements of electrons: 1s2, 2p2, 2p6 etc. and I understand how the configurations are derived from the quantum numbers. I am curious if the distribution of electrons can be explained in terms of electrostatic forces, i.e. mutal repulsion of electrons and attraction by the nucleus. That is, if you made a 3-d computer model of point charges around a larger central opposite charge, would you find stable configurations that correspond to anything like the electron configurations you get from the four quantum numbers? I suppose the spin number doesn't fit, but how about quantum numbers n, l, and m? I know the idea of electrons being point charges at determinate locations is completely replaced in quantum mechanics. I'm just wondering if the configurations (again, 1s2, 2s2, 2p6, etc.) are roughly explained as stable 3-d electrostatic arrangements of hypothetical point charges at specific points.

I hope this question makes sense. What I am really wondering is if atoms can be "explained away" in terms of more basic forces (i.e. electrostatic and strong nuclear), or is an atom more than the sum of its parts, that is, is it something you can't presently derive from more basic physics.

Thanks.
 

Answers and Replies

  • #2
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If you suppose that the nucleons and electrons only need to solve Maxwell's equations, and not the mathematics of quantum mechanics, you discover that even in the hydrogen atom (an electron-proton bound state) the electrostatic attraction implies that H is fundamentally unstable.

Quantizing fundamental properties of the H atom assures us that H is stable, as we expect.
 
  • #3
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the electrostatic attraction implies that H is fundamentally unstable.
What about the full electromagnetic interaction?

Quantizing fundamental properties of the H atom assures us that H is stable, as we expect.
If we imagine a macroscopic charged object orbiting a much heavier charged object, and doing so rather fast, then solving Maxwell's equations we'd get that the two objects break up?
 
  • #4
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TimH:
I think the answer to your question is: Yes, BUT you have to use the Schroedinger equation to do it properly. The atomic SE has two terms: one for kinetic energy and one for potential energy. The potential energy term is simply the 1/r function of Coulomb interation - that is electrostatics. Solutions to the SE are quantized and explain observed atomic line spectra, in addition to all other atomic properties. Another way to say this is that electrostatic interations real, but in atoms they must be treated quantum mechanically for accurate results. Classical models don't give quantization, aside from the obvious problem of electrons collapsing into nucleus.
Nonetheless, in the area of molecular physics there are two results that can be obtained from classical electrostatic considerations. First is a theorem known as the electrostatic theorem. It says that the forces on an atom in a molecule computed from quantum mechanics are identical to those one would obtain from a classical computation using the quantum electron distribution. The second is the VSEPR theory, which is used to rationalize and predict molecular geometries. It considers molecular geometries are those that minimize the repulsion between valence electron pairs. This theory is described in detail in most first year chemistry texts.
Jim Ritchie
 

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