Evidence of orbitals?

Question

How do we know that there are different types of orbitals? For example, what evidence is there for the existence of $\mathrm{p}$ orbitals instead of there being multiple $\mathrm{s}$ orbitals (for example, why isn't the electronic configuration of sodium $\mathrm{1s^1, 2s^2, 2s^2, 2s^2, 2s^2, 3s^2}$ instead of $\mathrm{1s^2 2s^2 2p^6 3p^1}$)?

http://en.wikipedia.org/wiki/Crystal_field_theory provides evidence for existance of orbitals with lower symmetries. — permeakra, Sep 12 '14 at 14:51

score 21 · Answer 1 · edited May 29 '19 at 16:04

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Let me approach this another way than the others: orbitals are NOT physical objects! They do not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn–Sham / etc. equations.

Orbitals are not unique: given linear combinations are equivalent with each other, and there is no "correct orbitals", one can choose whichever they like. Canonical orbitals, natural orbitals etc are all good to go.

What is the evidence they exist? They do not exist, they are just mathematical solutions for given equations, and it is a purely mathematical question if they are good solutions for those equations or not. The theories themselves are consistent with experimental data, e.g. spectroscopic properties, geometries, reactivity.

edited May 29 '19 at 16:04

Glorfindel

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answered Sep 12 '14 at 16:35

Greg

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The answer that I personally was looking for. – LordStryker Sep 12 '14 at 16:42
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@Greg wote, "They do not exist" Electrons exist and they occupy regions with probabilities defined by equations. These regions may have different spatial functions. "Orbital" is a word that can be used to describe that spatial function\distribution of electrons. In that sense of the word, orbitals exist, just as the spatial distribution of the electron exists. Perhaps this is more a semantics issue. – ron Sep 13 '14 at 01:39
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It is not semantic issue. Which one exist? Canonical ones? NBOs? Or some of their rotations? – Greg Sep 13 '14 at 02:52
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"given linear combinations are equivalent with each other" if we are talking about energy – user1420303 Sep 24 '16 at 12:46
Not all linear combinations are equivalent, only unitary transformations. 2. This also only holds true if we look at the overall energy of the system. If we were to assume that orbital energies themselves are physically meaningful, then this does not hold anymore. 3. The question whether orbital energies are in any way observable harks back to the old question about how to interpret the photoelectron spectrum of methane and its distinct 3+1 intensity pattern. To this day I still haven't found a satisfying answer.

Antimon

Nov 11 '19 at 05:02

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@Antimon 1) OK, clarifying: not all are equivalent, only an infinite subset of them... Still, it doesn’t make them neither unique, nor physical, only the solutions of one-electron given approximations. 2) Orbital energies are only meaningful as approximations. Also, which one? HF? DFT? 3) The question was for proof for orbitals. Just because orbital energies can be useful approximations, it doesn't mean they are proof. Two-body solutions for the Earth’s orbital are useful, but they are not a proof that only two body exist in the universe. – Greg Feb 21 '21 at 17:51

lagrange103 · Answer 2 · 2019-05-30T21:03:37.693

The answer lies with experimental chemistry, specifically successive ionisation energies (i.e how much energy is required to remove the first electron, the second electron, the third electron and so on).

Each point on the graph corresponds to an element. The first one is hydrogen, the second is helium. The height of each point shows how much energy is required to remove the first electron.

You can note that generally, the energy increases over a period. This is because In each successive element there is one more proton, and this stronger nuclear charge 'hold the outer electrons' more tightly.

Now to explain your question. Observe that within a period (e.g the 3rd dot to the 10th), it is not a constant increase. You can see that between the 4th-5th there is a slight drop, likewise between the 7th-8th.

The explanation for this is the sub orbitals. Some knowledge you need to know is that electron sub shells are only stable when empty, full, or half full (if you need an explanation for this comment on it later).

Let us examine the fourth dot, which represents Beryllium. It has an electronic config of 1s2 2s2. All of its sub orbitals are full, meaning that it is quite stable. Compare it to the 5th dot, boron. Boron has a config of 1s2 2s2 3p1. Now the P orbital has room for 6 electrons, but this only has 1! It is not happy. It is not full or half full. Because of this it is trying to 'get rid' of the electron to be more stable. That's why it doesn't require so much energy to remove the outer electron.

The decrease between the 6th and 7th is explained by the fact that the p orbital is stable when empty, full or half full. The 7th dot (nitrogen) has 3 electrons in its p orbital (half full). Contrasted with oxygen, which has 4/6. This is not stable, so less energy is required to remove it.

TL-DR: By analysing the ionisation energy graphs, we can see patterns that can be explained by sub orbitals.

If you need a more basic/complex explanation, comment.

I'd be a bit careful about the whole sub-shells being stable with empty, half-filled, and filled description. Slater's rules suggest these are general tendencies, but not the only story when you get into the transition metals and lanthanides/actinides. But that's a minor quibble. — Geoff Hutchison, Sep 12 '14 at 13:47

score 10 · Answer 3 · answered Sep 12 '14 at 13:45

Oh, I suspect someone could come up with a "s-orbital only" view of chemistry, much like people came up with complicated models for an Earth-centered view of the universe.

Let's start with the simple fact that solving quantum mechanics for a hydrogen atom gives you solutions for s, p, d, f, g.. orbitals, and even the degeneracy (i.e., that there is one s-type orbital, three p-type, five d-type, etc.) based on the angular momentum.

As described above, solutions for the many-electron equations (albeit approximate) match up very nicely with experimental observations for ionization energies, electron affinities, etc.

So an elegant theory and experiment agree to a remarkable degree.

Beyond that, we know that there must be non-spherical orbitals because we see molecular shapes with bond angles. I cannot come up with any way to describe a tetrahedral methane (let alone anything else) without some sort of non-spherical orbital.

Moreover, when we look at reactivities of molecules, we see reactions occurring where we predict lone pairs or radical spin-density, etc.

While orbitals are truly a mathematical construct, we find they are incredibly predictive of a wide range of chemistry. So my response would be "how can we think there are not different kinds of orbitals?"

Given the simplicity of s orbitals, people have tried using 's-orbital only' basis sets and just placing a crap ton of these orbitals all over a molecular space rather than using a smaller number of more 'accurate' atomic orbitals. Interesting concept. — LordStryker, Sep 12 '14 at 14:01
Minor quibble - the tetrahedron of methane could easily be described as resulting from a spherical C^4- surrounded by four point positive charges. The tetrahedron optimizes the repulsion of the positive charges from each other while keeping them all at the same distance. Not a criticism of your answer in general, just the choice of methane as the example that requires nonspherical orbitals. — Andrew, Feb 21 '21 at 14:04
Technically you can describe everything with sin/cos functions, too, yet it doesn't mean that the vacuum has tons of orbitals that look like a sic/cos function. Let’s distinguish model from physical reality, especially in places where the model is strictly for tied to an approximation. — Greg, Feb 21 '21 at 17:56

score 9 · Answer 4 · edited Feb 21 '21 at 12:35

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While I really like the other answers, the 21th century evidence here is that we have images of these orbitals thanks to atomic force microscopy (AFM).

http://www.themarysue.com/electron-orbital-pictures/

As getafix pointed out, these are actually images of spatial electron density distributions by the examined molecules. The actual evidence here, that the model which uses the orbital term predicts the same distributions as we measured with AFM.

edited Feb 21 '21 at 12:35

Alchimista

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answered Sep 24 '16 at 20:44

inf3rno

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What you evidence of orbitals, can just as easily be called evidence of regions of electron density. ;) So is orbital a real thing or just a description of the spatial distribution of electrons. Electrons are real, regions with a high probability of finding them are real, but are orbitals? food for thought.. – getafix Sep 25 '16 at 01:30
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@getafix Nice thoughts. :-) This depends on the definition of orbitals. So yes these images are evidence that there is a spatial electron density distribution and that distribution is very similar to the prediction of the limited mathematical model which uses this orbital term. Usually this is enough to prove a hypothesis. E.g. by general relativity such a prove was that by a solar eclipse they were able to check how the gravity of the Sun bends the space and so starlight... – inf3rno Sep 25 '16 at 02:59
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@getafix But yes you are right, I could rephrase the text. So these are not images of orbitals, but images of electron density distributions which are the same as the model predicts. – inf3rno Sep 25 '16 at 03:01
Well is also somehow similar to fields. While iron filings gives directly the field lines, and in this case more steps are needed, in both case we need a physical handle to picture the underlying thing. Is much of semantic. – Alchimista Feb 21 '21 at 12:53
Not even electron distribution: only a slice of those close to the Fermi level. – Greg Feb 21 '21 at 17:58

score 7 · Answer 5 · edited Sep 24 '16 at 03:35

How do we know that there are different types of orbitals?

In the early days of atomic spectroscopy scientists were able to explain the spectral properties of elements like hydrogen and sodium. Atoms of these elements had a single electron in the outermost shell and produced spectra that fit with (what we now view as) the relatively simplistic Bohr theory of the atom. When more complex atoms were examined spectroscopically, fine structure was observed that the Bohr theory could not explain. These spectral lines were often referred to as sharp, principal, diffuse and fundamental.

As quantum mechanics emerged, the concept of the 4 quantum numbers being needed to describe an electron surfaced. It was found that the spectral fine structure could now be explained when these various quantum states were taken into account. To honor the work done by the early spectroscopists (that led to the development of a better theory), scientists used their "s, p, d, f" notation to describe the various values for the angular momentum quantum number $\ell$.

So although the observations made by the early spectroscopists couldn't be explained at the time, it turns out that they were actually observing spectral transitions involving other orbitals (p, d, f) in addition to those involving the s orbital. The observation of this spectral fine structure requires the presence of orbitals that are different from s orbitals.

Evidence of orbitals?

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