32
$\begingroup$

The Bohr model of the atom is essentially that the nucleus is a ball and the electrons are balls orbiting the nucleus in a rigid orbit.

enter image description here

This allowed for chemists to find a model of chemical bonding where the electrons in the outer orbits could be exchanged. And it works pretty well as seen in the Lewis structures:

enter image description here

However, electron orbitals were found to be less rigid and instead be fuzzy fields which, instead of being discrete/rigid orbits, look more like:

enter image description here enter image description here

However, in chemistry education like organic chemistry you still learn about chemical reactions using essentially diagrams that are modified lewis structures that take into account information about electron orbitals:

enter image description here

What I'm wondering is, if the Bohr model is used essentially throughout college education in the form of these diagrams, it seems like it must be a pretty accurate model, even though it turns out atoms are more fuzzy structures than discrete billiard balls. So I'm wondering what the inaccuracies are, and if there is a better way to understand them than the Bohr model. If you build a computer simulation of atoms with the Bohr model, I'm wondering if it would be "accurate" in the sense of modeling atomic phenomena, or is it not a good model to perform simulations on. If not, I'm wondering what an alternative model is that is better for simulation. Essentially, how good the Bohr model is as a diagram, as a tool for learning, and as a tool for simulation.

$\endgroup$
  • 1
    $\begingroup$ Don't forget about the little problem of having zero angular momentum in an orbit! $\endgroup$ – JDługosz Mar 27 '18 at 1:09
  • 3
    $\begingroup$ What is the part with 1s, 2p, 3d, 4f and the other related "diagrams" illustrating? I understand the other graphics in your question just not this. $\endgroup$ – d-b Mar 28 '18 at 9:29
  • 1
    $\begingroup$ That's the electron orbital/configuration en.wikipedia.org/wiki/Electron_configuration. $\endgroup$ – Lance Pollard Mar 28 '18 at 18:55
  • 1
    $\begingroup$ Biggest problems: In the ground state the electron is accelerating but doesn't generate electromagnetic radiation? Excited states don't decay to the ground state? $\endgroup$ – jim Mar 28 '18 at 19:37
  • 3
    $\begingroup$ You make a lot of statements that the Bohr model has various successes, but basically every statement you make is false. The Bohr model gets essentially nothing right other than the energy levels. $\endgroup$ – Ben Crowell Mar 28 '18 at 19:43
42
$\begingroup$

In hydrogen:

  1. It incorrectly predicts the number of states with given energy. This number can be seen through Zeeman splitting. In particular, it doesn't have the right angular momentum quantum numbers for each energy levels. Most obvious is the ground state, with has $\ell=0$ in Schrodinger's theory but $\ell=1$ in Bohr's theory.
  2. It doesn't hold well under perturbation theory. In particular, because of angular momentum degeneracies, the spin-orbit interaction is incorrect.
  3. It predicts a single "radius" for the electron rather than a probability density for the position of the electron.

What it does do well:

a. Correct energy spectrum for hydrogen (although completely wrong even for helium). In particular, one deduces the right value of the Rydberg constant.

b. The Bohr radii for various energy levels turn out to be the most probable values predicted by the Schrodinger solutions.

c. Also does a lot of chemistry stuff quite well (as suggested in the original question) but I'm not a chemist so can't praise the model for that.

$\endgroup$
  • 6
    $\begingroup$ The point about the number of states with a given energy is a good one. In particular, note that the "magic number" of eight electrons in a Lewis diagram relies on the $n = 2$ orbitals being able to accept eight electrons. This is not a prediction of the Bohr model, and it's hard to see how to modify it to account for this. $\endgroup$ – Michael Seifert Mar 26 '18 at 17:40
  • 1
    $\begingroup$ Good information here, but to say that it predicts any chemistry seems a colossal stretch. It has no mechanism for bonding at all. $\endgroup$ – DWin Mar 27 '18 at 1:27
  • $\begingroup$ @DWin The chemists use it a lot so there must be some roughly right about it. But then again, I’m not a chemist... $\endgroup$ – ZeroTheHero Mar 27 '18 at 2:08
  • 8
    $\begingroup$ I'm not a chemist either but my experience with chemistry taught in the latter half of the 20th century was that we were taught about electron clouds and pairing of outer shell electrons. Electron pairing and hybrid orbitals do not seem to be an immediate consequence of the Bohr model. I rather argue that chemists use a version of Pauling and Wilson's "Introduction to Quantum Mechanics with Applications to Chemistry" (c) 1935, 1963. My copy is the Dover edition. $\endgroup$ – DWin Mar 27 '18 at 3:56
16
$\begingroup$

This is an example of the "correspondence principle" in the broadest sense, that new theories should explain why old ones got some things right. The linked article discusses the Bohr model, but leaves some of your sub-questions unanswered. Going beyond that, how does an "electrons are somewhere specific" approximation lead to useful models of sharing and transferring electrons in covalent, ionic and metallic bonding? Well, we'll focus on covalent for now.

When physicists teach undergraduates enough quantum mechanics to do the hydrogen atom properly, electrons end up in specific atomic orbitals due to their quantum numbers, and each orbital can hold at most 2 electrons. The applications of Bohr-like reasoning you've brought up concern molecular orbitals, and these are a slightly more advanced topic; at this point I wish I knew what chemistry undergrads are taught about them, but I imagine Peter Atkins explains MOs with much the same rigour.

Like atomic orbitals, $\pi$ MOs hold at most 2 electrons (let's not get into $\sigma$ bonding for the moment). The Bohr lie would be that the electrons living in these orbitals have a precise location, and that orbitals form so as to get the electron count in each atom's outermost shell right and make for a stable molecule - you know, the usual $8$-electron rule (or $2$ for hydrogen, since it's trying to be like helium, not neon). The short answer to your question is that when we transition from quantum numbers for electrons in monatomic allotropes of an element to the analogous treatment of a molecule, the way the pattern of legal orbitals transforms is the same as would be expected on a classical model. Why? Because all you really need is the legal-number-combinations rules, not the way it's derived from the Schrödinger equation.

Let's consider the simplest possible example, $\mathrm{H}_2$. The simple model says, "we have one legal orbital, and it has room for $2$ electrons, which is just what we need, and they end up in an orbit like planets in a binary-star system". The more accurate model is, "again we fill the unique legal orbital with $2$ electrons, but the electrons' behaviour is quantum-mechanical". You can approximate the electrons in that orbital as two particles in a box (although that's not a perfect analogy), because they don't have enough energy to escape unless a photon excites them, nor can they fall into a lower orbital because those are full. With this constraint, the quantum effects are quantitative but don't make much of a qualitative difference.

$\endgroup$
  • $\begingroup$ not sure what you mean by " legal-number-combinations rules" $\endgroup$ – Lance Pollard Mar 26 '18 at 17:46
  • $\begingroup$ @LancePollard In the monatomic case $n\ge 1,\,0\le\ell\le n-1,\,-\ell\le m\le \ell,\,s=\pm\frac{1}{2}$. $\endgroup$ – J.G. Mar 26 '18 at 18:01
  • 1
    $\begingroup$ "legal combinations" of electron energy states i.e. the rules that lead to Lewis structure diagrams. $\endgroup$ – JimmyJames Mar 26 '18 at 18:07
  • $\begingroup$ I have some questions from a layman: Say you can calculate the probability of finding an electron somewhere, and it has a high probability in a certain cloud of space, Is there a > 0 probability of finding that electron anywhere in the Observable Universe? If yes, is this at all related to "virtual particles" and how so? Does the probability falloff exponentially with distance or what? $\endgroup$ – George Mar 27 '18 at 19:25
  • $\begingroup$ @Nofacr Yes (unless it's in a "box" but that's unphysical), no, yes or something like that depending on the wave function. $\endgroup$ – J.G. Mar 27 '18 at 19:46
14
$\begingroup$

The parallel between the Bohr picture and the Lewis diagrams isn't that great if you consider that the electron is moving in the Bohr model, while the electrons are static in a Lewis diagram.

If a Bohr electron was "at rest" outside a nucleus, as it is in a Lewis diagram or one of your organic-chemistry diagrams, it would immediately accelerate towards the nucleus. And I cannot see how you would modify a Lewis diagram so that the electrons were "shared" while still being in orbit around the nuclei.

$\endgroup$
11
$\begingroup$

Lewis dots and molecular structure diagrams, as a practical notation used by chemists, have very little to do with the Bohr model of the atom. They capture a set of empirical, qualitative rules which are true for most atoms under most conditions, and which were discovered first:

  1. Every atom has some number of "valence electrons" which control how it can combine with other atoms. The number usually ranges from zero to eight. The number of valence electrons possessed by an isolated atom with no electric charge is determined by which element it is.
  2. Atoms can have a different number of valence electrons from the default for that element, but then they will also have an electric charge, and we call them "ions".
  3. Chemical compounds are usually most stable when each atom of the compound can be said to have either zero or eight valence electrons (or, in one very important special case, namely hydrogen, two valence electrons).
  4. Atoms can combine to fulfill rule 3 in two ways. They can fully transfer electrons from one to another, forming ions, and then be held together by electrostatic attraction. Or they can "share" pairs of electrons between two atoms, in which case both electrons contribute to the valence electron count of both atoms, and then be held together ... well, ultimately again by electrostatic attraction, but a more targeted variety, such that we can say that each atom is "covalently bonded" to specific other atoms; this is not the case for ions.
  5. Combining rules 1, 3, and 4 lets you predict how many covalent bonds an atom of a particular element can form. For instance, carbon has four valence electrons in its uncharged, unbonded state, each of which can be shared with one other atom, so it can make up to four covalent bonds. Chlorine, on the other hand, has seven valence electrons, which means it can form only one covalent bond (any more and it would have too many valence electrons).
  6. Two atoms can share more than one pair of electrons, which we call a "double bond" or "triple bond" (four is very rare, five would break the 'usually zero to eight' rule -- I'm not going to say impossible but I've never heard of it); these bonds are physically stronger but also more reactive than single bonds.
  7. If you have a double bond next to a single bond and the overall molecule is sufficiently symmetric, what actually happens is the second bond gets "delocalized" over all three atoms. This can keep happening down a chain of alternating single and double bonds, and sometimes makes the molecule extra stable (e.g. benzene) or changes what chemical reactions the molecule will undergo (e.g. enols).

The Bohr model doesn't try to predict most of this; it is only concerned with isolated atoms. Bohr probably did have rules 1, 2, and 3 in mind when he developed it, though.

Schroedinger's equation isn't a model of atoms at all, it's the quantum equivalent of $\mathbf{F}=m\mathbf{a}$; to predict anything with it you have to define what the forces are. That's the domain of atomic and molecular orbital theory, and the actual quantitative Hamiltonians get really messy really fast; recovering the above rules for systems of more than about two atoms was still cutting-edge theory as of when I decided I didn't want to do that for a living (admittedly, that was 15 years ago now).

$\endgroup$
  • 1
    $\begingroup$ 5 and even 6 bonds are possible. The "8 electron' rule comes from light elements with just 1 s and 3 p orbitals, but dimolybdenum and ditungsten are formed from very heavy elements. E.g. in molybdenum the electrons participating come from the 4d, 5s and 5p orbitals. $\endgroup$ – MSalters Mar 28 '18 at 10:40
6
$\begingroup$

I can remember the course on introduction in quantum physisc when we discussed the Bohr model. The reasoning was this:

  • Electron circulating around the core on circular trajectory has nonzero acceleration.
  • Therefore it must emit radiation.
  • Therefore it loses energy. Therefore it must either "get a punch from somewhere" or fall in deeper and deeper orbit levels.
  • Finally it must fall in the core.

But, we do not observe such orbital decay and radiation emission, so the model is flawed.

On the other hand, Bohr's model is easy to imagine and there are nice parallels between "our big universe" and the "tiny universe down there". Once you accept the electrons are following strictly defined trajectories with corresponding energies, it is easier to accept that the electrons are somewhere in a more loosely define volume with energies still strictly defined. So. we introduce the word "orbital" to replace "orbit" to make a distinction between the probability volume of an "orbital", versus the exact radius implied by the word "orbit".

$\endgroup$
3
$\begingroup$

Lewis dot structures were not based upon the Bohr model.

Oppositely, upon introducing the dot structures in The Atom and the Molecule, Lewis specifically states:

Bohr in his electron moving in a fixed orbit, have invented systems containing electrons of which the motion produces no effect upon external charges. Now this is not only inconsistent with the accepted laws of electromagnetics but, I may add, is logically objectionable, for that state of motion which produces no physical effect whatsoever may better be called a state of rest.

...

I believe that there is one part of Bohr’s theory for which the assumption of the orbital electron is not necessary, since it may be translated directly into the terms of the present theory. He explains the spectral series of hydrogen by assuming that an electron can move freely in any one of a series of orbits in which the velocities differ by steps, these steps being simply expressed in terms of ultimate units (in his theory Planck’s h is such a unit), and that radiation occurs when the electron passes from one orbital velocity to the next. It seems to me far simpler to assume that an electron may be held in the atom in stable equilibrium in a series of different positions, each of which having definite constraints, corresponds to a definite frequency of the electron, the intervals between the constraints in successive positions being simply expressible in terms of ultimate rational units

...

the most stable condition for the atomic shell is the one in which eight electrons are held at the corners of a cube

So, for example, diatomic iodine was considered by Lewis as two cubes sharing an edge.

$\endgroup$
  • $\begingroup$ nice contribution! $\endgroup$ – ZeroTheHero Mar 28 '18 at 19:46

protected by Qmechanic Mar 26 '18 at 19:47

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.