As you all know, iodine has a purple color when in the vapor phase and when dissolved in non-coordinating solvents such as carbon tetrachloride. What electronic transition is responsible for this?

A quick google search (https://www.vedantu.com/question-answer/the-color-of-iodine-is-a-violet-b-green-c-white-class-11-chemistry-cbse-5f4a918f84fe9e109c4b5753) says that this electron transition is due to non-bonding electrons going into the antibonding sigma orbital.

This is certainly possible, but leads me to a more general question: Usually, we don’t draw the non-bonding orbitals into the MO-scheme of the molecule. Take a look at the fluorine molecule for example (as its MO scheme should be similar than that of iodine):

enter image description here

Fluorine also has lone pairs, and they also could transition electronically; why don’t we write them out specifically and how does one know when they will be involved in electronic transitions?

For example, in carbonyls, there is also the n-pi* transition for reference,in which the non-bonding electrons of oxygen can be promoted to an antibonding orbital. Why don’t they participate in the MO scheme?

  • 2
    $\begingroup$ Electrons from lower orbitals (higher bonding energy) will transition to higher orbitals (lower bonding energy) when excited (by whatever...). Even forbidden transitions occur because atoms (or molecules) in the real world aren't totally isolated. Thus microstate perturbations allow "forbidden transitions," but they are relatively low in probability. The rub here is not all of the transitions (excitation or emission) occur in the visible band. $\endgroup$
    – MaxW
    Commented Aug 17, 2023 at 3:43
  • $\begingroup$ I'm not sure I understand the question. Lone pairs and electrons in non-bonding orbitals are not the same thing $\endgroup$
    – Ian Bush
    Commented Aug 17, 2023 at 7:56
  • $\begingroup$ @IanBush how come it is not the same? For example in carbonyls, the oxygen electron pairs are the cause of the non-bonding-pi* transition: masterorganicchemistry.com/2016/09/26/… $\endgroup$
    – Mäßige
    Commented Aug 17, 2023 at 12:38
  • 1
    $\begingroup$ I2 has the ground state configuration $(\sigma_g5p)^2(\pi_u5p)^4(\pi^*_g5p)^4(\sigma^*_u5p)^0$ and the first excited state $(\sigma_g5p)^2(\pi_u5p)^4(\pi^*_g5p)^3(\sigma^*_u5p)^1$, so $\pi^* \to \sigma^*$. The transition is between states $^1\Sigma^+_g \,\to\,^3\Pi_u$ . As this molecule contains heavy atoms spin-orbit coupling is very important and the singlet - triplet spin transition prohibition is broken. $\endgroup$
    – porphyrin
    Commented Aug 17, 2023 at 16:15
  • 2
    $\begingroup$ To be sure I would have to read up on this more, but my first guess would be as you say, pi* -> sigma* But the point @porphyrin makes about spin orbit coupling is important as well $\endgroup$
    – Ian Bush
    Commented Aug 17, 2023 at 19:00

1 Answer 1


Hybrid orbitals and molecular orbitals are different ways of looking at the same thing

Let’s take your fluorine example first.

It doesn’t quite make sense to ask why we “don’t write out the lone pairs specifically” in $\ce{F2}$; we’re using a different model. Once we’ve decided that the $1s$ electrons are core and won’t participate significantly in bonding, we have 14 electrons to worry about, all of which are shown in the MO diagram; there are no extra lone pairs that we’ve chosen to ignore.

Alternatively (but worse from the perspective of understanding electronic transitions), we could say that these 14 electrons are arranged in three lone pairs on each of two F atoms (3 pairs $\times$ 2 atoms $\times$ 2 electrons per pair = 12 electrons total) plus one $\sigma$ bond between the atoms (the final two electrons).

So it’s not that we’re choosing to ignore lone pairs, it’s that they don’t exist in this model.

Non-bonding orbitals are typically included in MO diagrams

The link you gave to illustrate your carbonyl example only shows three orbitals. For a fuller picture, let’s have a look at a qualitative MO diagram for the simplest carbonyl compound, formaldehyde:

Qualitative MO diagram for H2CO, shown in terms of interactions between CO and H2

(This diagram has some flaws but it was the best I could find quickly online.)

The LUMO, here given the symmetry label $2b_1$, is an example of an orbital conventionally labelled “non-bonding”, though in fact you can see that it is mildly antibonding between C and O and bonding between C and H. There’s a more quantitative picture here. The point is that we don’t have to decide whether or not to include this; it turns up naturally when we make the diagram.

The visible transition in $\ce{I2}$ is better thought of as being between two antibonding orbitals

Returning to your initial question, as your own MO diagram and several comments have indicated, the HOMO-LUMO transition in $\ce{I2}$ is best described as $\pi^*\rightarrow\sigma^*$, not $n\rightarrow\sigma^*$. The source you link to for the latter description is not very scholarly! As the comments have pointed out, the reason for iodine’s strong colour includes both that this transition occurs in the visible spectrum and that spin-orbit coupling breaks the “selection rule” prohibiting a singlet-triplet transition.


Your Answer

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

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