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It is known that the halogens have the following colours:

$\ce{F_2}$: Pale Yellow
$\ce{Cl_2}$: Greenish Yellow
$\ce{Br_2}$: Reddish Brown
$\ce{I_2}$: Violet

If we talk about predicting the colours based on their HOMO-LUMO transition energies, we may say that the $π^*_{np}$ - $σ^*_{np}$ gap decreases with increasing $n$, thus the wavelength of energy absorbed should increase and correspondingly the wavelength of emitted light should decrease as we go from $\ce{F_2}$ to $\ce{I_2}$.

However, while $\ce{F_2}$, $\ce{Cl_2}$ and $\ce{I_2}$ seem to adhere to this order, $\ce{Br_2}$ doesn't, showing up on the far end of the spectrum.

What is the reason behind this anomaly? Here's an illustration: enter image description here

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  • $\begingroup$ Are you asking about the elements in the gas phase, or for the elements in their standard states? $\endgroup$
    – Ian Bush
    Commented Aug 10 at 7:57
  • $\begingroup$ Standard states, I suppose. Does the phase matter though? Not sure but it seems counterintuitive for a substance to change colours just through phase change. $\endgroup$
    – Eren Jäger
    Commented Aug 10 at 8:18
  • $\begingroup$ Furthermore, if there isn't any difference in the electronic configurations in different phases, how could there be different colours? Pardon me if I am unaware of other factors at play here as I'm only a high school student $\endgroup$
    – Eren Jäger
    Commented Aug 10 at 8:23
  • $\begingroup$ Look at the pictures of Iodine on wikipedia - the solid is definitely a different colour from the gas. Put simply in the rarefied gaseous case you can consider individual molecules as they are so well separated from each other, while in the much more dense condensed liquid and solid phases interactions between molecules become important, so much so in the iodine case that it can be considered a semiconductor. $\endgroup$
    – Ian Bush
    Commented Aug 10 at 9:40
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/33861/… $\endgroup$
    – Nilay Ghosh
    Commented Aug 10 at 9:52

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Anomaly is the language of chemistry !

The color of halogens—from fluorine’s pale yellow to iodine’s violet —reflects their electronic transitions, specifically the HOMO-LUMO gap, which narrows as we move down the group. This leads to absorption of longer wavelengths, explaining the progression in color. However, bromine stands out reddish brown hue, deviating from the expected trend. This anomaly arises due to spin-orbit coupling, which significantly alters its electronic energy levels, and its unique molecular interaction (like electron correlation effects and intermolecular interactions). These factors shift bromine’s absorption spectrum, resulting in its distinctive color, highlighting the complexity of predicting such properties

Edit : Iodine's stronger spin-orbit coupling pushes its absorption further into the visible spectrum, causing it to appear violet, consistent with the expected trend. In iodine, the much stronger spin-orbit coupling causes a more significant splitting of energy levels. This shifts the absorption into the higher wavelength (lower energy) part of the visible spectrum, making iodine appear violet.

The spin-orbit coupling in bromine is moderate. It slightly splits the molecular energy levels, causing bromine to absorb light in the blue-green region.

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  • 1
    $\begingroup$ Spin-orbit coupling will be even stronger in iodine than bromine. Personally I suspect that the change of phase from gas to liquid to solid has at least as much influence as any molecular changes, but without a reference to support me I won't simply assert it. $\endgroup$
    – Ian Bush
    Commented Aug 10 at 7:20
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    $\begingroup$ The order of electronic energy gaps is $I_2\lt Br_2\lt Cl_2 \lt F_2$ , so spin -orbit effect do lower each of the heavy atom's spectrum, but the way we perceive color depends on all absorption wavelengths in some complicated way. $\endgroup$
    – porphyrin
    Commented Aug 10 at 7:27
  • $\begingroup$ Unknown - You might not be aware but the person whose edit you accepted happens to be a troll: chemistry.meta.stackexchange.com/questions/5404/… and chemistry.meta.stackexchange.com/questions/5399/… $\endgroup$
    – Nilay Ghosh
    Commented Aug 10 at 14:33
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    $\begingroup$ @Unknown Hi, I think that's the best possible way to explain it, my teacher too said somewhat the same thing though I didn't understand it back then.. Although that seems like a rather empirical explanation than theoretical $\endgroup$
    – Eren Jäger
    Commented Aug 10 at 20:25
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    $\begingroup$ Please use bold, italics and uppercase sparingly. $\endgroup$
    – Buck Thorn ♦
    Commented Aug 11 at 8:13

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