According to Wikipedia, HF is formed by the interaction between hydrogen's $1\mathrm{s}$ orbital and fluoride’s $2\mathrm{p}_z$ orbital:

In hydrogen fluoride HF overlap between the H 1s and F 2s orbitals is allowed by symmetry but the difference in energy between the two atomic orbitals prevents them from interacting to create a molecular orbital. Overlap between the H 1s and F 2pz orbitals is also symmetry allowed, and these two atomic orbitals have a small energy separation. Thus, they interact, leading to creation of σ and σ* MOs and a molecule with a bond order of 1.


  1. If there is a too large energy separation between H-1s and F-2s, isn't the energy separation between H-1s and F-2p even larger?
  2. Can it be experimentally verified that the bonding is indeed via the F-2p and not the F-2s?
  • 3
    $\begingroup$ No. The F2p is much closer in energy to the H1s than the F2s is. You have to remember that fluorine is a highly electronegative element and as a result all the orbital energies are lower than that in hydrogen. $\endgroup$ Jan 18, 2016 at 7:12
  • 1
    $\begingroup$ @orthocresol why high electronegativity results in lower energy orbitals? $\endgroup$
    – Sparkler
    Jan 18, 2016 at 13:48
  • $\begingroup$ Is the lower energy of highly electronegative element due to the closeness of the orbital to the nucleus? $\endgroup$ Jan 24, 2021 at 17:06

3 Answers 3


You fell into a wrong analogy with the picture of energetically far apart. Let me break down your picture into something visible:

Imagine the orbitals of fluorine at different heights of a mountain (I’ll use the Zugspitze, because it feels like home). 1s is at the very peak of the mountain (yeah, I’m turning this around just for simplicity). The Eibsee (c. $1000~\mathrm{m}$ over sea level) would correspond to the 2s-orbital. Garmisch-Partenkirchen, the town in the valley before the mountain, ($700~\mathrm{m}$) would be the 2p orbital.

We need to place the 1s-orbital of hydrogen somewhere so that the height difference to the Eibsee is too large for an interaction. One possibility would be the Höllentalanger hut at around $1300~\mathrm{m}$. That would render the energy difference to both fluorine’s 2s and 2p too large. But in fact, hydrogen’s 1s is better placed at the height of Oberau, $650~\mathrm{m}$. That is way too far away from the 2s but rather close to the 2p.

This is confirmed by energy diagrams of $\ce{HF}$ like that presented in this lecture (in German). (Just search for HF on the page, it’s not far down. Ignore the text around it and look at the scheme.)

Now you may be asking yourself why the hydrogen’s 1s energy is so high compared to fluorine’s orbitals. Compare ionisation energies, which are typically understood as the energy difference of the highest orbital to ‘an electron in vacuum’. For hydrogen, this energy is $1312~\mathrm{kJ/mol}$ while for fluorine this is $1681~\mathrm{kJ/mol}$. The reasoning behind it is fluorine’s very high electronegativity, attracting the electrons towards its nucleus. Fluorine’s nucleus has a $+9$ charge which is much better at drawing electrons in than hydrogen’s $+1$ (even though it needs to draw nine electrons in rather than one). This stabilises all of fluorine’s orbitals; and those which are ‘closer to the nucleus’ (s-orbitals) most.

I am not well-educated in modern physical chemistry methods, but I heard that orbitals were visualised. If that is the case, it won’t be long until we can visualise the $\ce{HF}$ orbitals, and indeed show that the bonding orbital (highest energy σ type) will have a nodal plane through fluorine.

  1. No, there is larger energy separation between H$_{1s}$ and F$_{2s}$. To see this lets lie a lot and say that orbital energies are proper energies. So thinking classically, the energy is due to coulomb interaction between the nuclei and the electron, electrons always have the same charge, but fluorine nuclei have much more charge than hydrogen. That means that (in this picture) the electron in a H$_{1s}$ would have greater energy than one in F$_{1s}$ (the second one much greater in absolute value). An F$_{2p}$ electron "feels" like there is not as much charge in the nuclei (because the coulumb repulsion with inner electrons), so its energy is closer to the H$_{1s}$ energy.

By the way, I have to say it, I do not find much sense in the argument about the closeness of orbital energy to assert about the overlaps.

  1. No, it can't be measured because orbitals does not exist. All we can do is perform molecular calculations to get an approximation to the wave function. Then force the separation of this function in hydrogen like orbitals (other functions with whom we are familiar or obsessed) through projections to make this kind of interpretations.
  • $\begingroup$ I'm fairly sure there is a way to show that the interaction between two AOs depends inversely on the energy difference - I think it might be done via perturbation theory but I don't know the details. It's probably in Albright, Burdett, & Whangbo's Orbital Interactions in Chemistry somewhere. $\endgroup$ Mar 2, 2016 at 22:59
  • $\begingroup$ @orthocresol I bet you are referring to a Møller–Plesset like treatment for the correlation energy (take a look at en.wikipedia.org/wiki/Møller–Plesset_perturbation_theory). But I'm not sure about what you mean by "interaction between two AOs". As they are abstract mathematical construction strictly speaking they can not interact. Electrons can interact and in the framework used in the question, I think that spatial overlap is important. Consider a molecule where similar orbitals are separated by many bonds, we won't use them to generate a molecular orbital. $\endgroup$ Mar 2, 2016 at 23:42
  • 1
    $\begingroup$ If you state something like "let's lie a lot" I kindly ask you to state exactly what you are lying about. In this case, why are orbital energies not proper energies. I guess you will state that orbitals do not exist, but I guess it is necessary that you back up this statement. As it stands here, it reads as just an opinion. You also must explain why you are using a model that you believe is wrong. $\endgroup$ May 2, 2016 at 7:41
  • $\begingroup$ @Martin-マーチン, My assertion that these are not proper energies is due that they are not eigenvalues of Hamiltonian of the system, but eigenvalue of the Fock operator. I did not add this kind of comments in the answer I thought that it was far too technical when considering the question. $\endgroup$ May 2, 2016 at 20:37
  • $\begingroup$ @Martin-マーチン , I just try to go along the question. The usage of this kind of mindset are sometimes useful to solve some problems. The "why" they are useful (considering that they are wrong), that is, why they turn to be (sometimes) acceptable representation of reality is many times harder than learning more more advanced descriptions. Note that I do not use the word "model" because they really aren't, as they are not clearly established. $\endgroup$ May 2, 2016 at 20:41

Wrong way round- search "orbital potential energy" and you'll know what I mean.

Hydrogen 1s is much higher in energy than F 2s and slightly higher then F 2p; that's why the bonding MO is centred mostly on fluorine and the unfilled LUMO is mostly on hydrogen.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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