Bromine ($\ce{Br2}$) has a dark reddish color. If it reacts with something like benzene, this results in the formation of Bromobenzene and $\ce{HBr}$:

$$\ce{Br2 + C6H6 ->[{Cat.}]C6H5Br + HBr}$$

It can be observed that during this reaction, the color of bromine disappears.

One can thus conclude that $\ce{HBr}$ is colorless. Why is this the case? If we look at the molecules, it seems to me that they are basically the same (speaking in how many electrons there are in each orbital, for example), except for that in $\ce{HBr}$, there is a hydrogen atom instead of another $\ce{Br}$.

So what makes the $\ce{Br2}$ bond "special", so that this molecule can absorb light in the visible spectrum but $\ce{HBr}$ can't?

I think that I have a basic understanding of the quantum mechanic model of an atom (e.g. atomic orbitals and how their energy levels correspond to different wavelengths of light being absorbed). Here are some of my ideas trying to explain the above described question:

  • In $\ce{Br2}$, both atoms have the same electronegativity, while in $\ce{HBr}$, there is a difference of $\ce{\Delta EN=0.76}$. The only consequence I could imagine from that is that in $\ce{HBr}$, the probability of an electron in the bond to be located closer to the $\ce{Br}$ atom is higher than being close to $\ce{H}$, while in $\ce{Br2}$, the electrons (or rather their probability) is evenly distributed

  • $\ce{H}$ has an electron configuration $\ce{[1s^1]}$, $\ce{Br}$ has an electron configuration $\ce{[1s^2]\,[2s^2]\,[2p^6]\,[3s^2]\,[3p^6]\,[3d^{10}]\,[4s^2]\,[4p^5]}$. So I would assume that the bond in $\ce{HBr}$ is formed by the overlapping of a $\ce{1s}$ and a $\ce{4p}$ orbital, while in $\ce{Br2}$, two $\ce{4p}$ orbitals overlap. The $\ce{4p}$ orbital has a higher energy level than $\ce{1s}$, so this might be the cause for the different absorption spectrum. However, I am not sure if this is true and how the energy levels behave when two orbitals overlap to form a new one.

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    $\begingroup$ The neutral Br atom has no electronic transitions in the visible range. The Br${2}$ molecule has them, because of vibrational states. $\endgroup$
    – Jon Custer
    Dec 9, 2020 at 19:15
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    $\begingroup$ Br is not a thing at all. Color is a function of molecules and other aggregates, and not of atoms. That's just the way it is. Nearly all elements are found in variously colored compounds, and many are found in colorless compounds. $\endgroup$ Dec 9, 2020 at 19:16
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    $\begingroup$ @Ivan, So you suggest there is no single molecule absorption/emission spectroscopy? It is probably a limitation of our eyes rather not a problem of atoms or molecules. $\endgroup$
    – AChem
    Dec 9, 2020 at 20:10
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    $\begingroup$ Why was this question closed? It is clearly not a homework question and I did share my thoughts $\endgroup$
    – jng224
    Dec 9, 2020 at 21:35
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    $\begingroup$ 1) We see something colorful if there is a permitted electronic transition between a ground state and an excited state corresponding to a wavelength of about 200 to 800 nm. 2) Isolated atoms' electronic transitions do not match this range, so they appear colorless to our eyes. 3) If atoms join to yield molecules, conceptually speaking, atom orbitals may mix to yield molecular orbitals. 4) Among these molecular orbitals, there may be permitted electronic transitions corresponding to a wavelength in the visible range; which depends on the atoms contributing. So $\ce{Br}$ is not $\ce{Br2}$. $\endgroup$
    – Buttonwood
    Dec 9, 2020 at 22:12

1 Answer 1


The colour of molecules come from electronic transitions that are very sensitive to the nature of the molecule

To simplify a little, molecules absorb light when the energy difference between electrons in an occupied orbital and an unoccupied orbital match the energy of a photon of light (and that light will have a specific colour). This much you already understand.

But the energy levels of the electrons that matter for colour are usually the bonds in the molecule that join one atom to another, not the other electrons in the atom that don't get involved in the bonding. So atoms are rarely thought of as having an inherent colour that is independent of what they are attached to (some isolated atoms do have colour: sodium vapour has narrow absorption bands in the yellow, for example, but this isn't important in sodium compounds).

The assumption you make that is false is that "If we look at the molecules, it seems to me that they are basically the same". If this were true there would be very little interesting in the whole subject of chemistry. When bromine reacts with benzene a Br-Br bond is replaced by a Br-C bond and a Br-H bond. Those are very, very different bonds. A Br-Br molecule has a readily accessible electron transition that results in a dark red colour. That this particular transition is very sensitive to the exact nature of the molecule is obvious as other dihalogens (with very similar electronic structures) have very different colours (iodine is dark purple, chlorine is light yellow). But the orbitals available in other Br-X bonds will be very different as will their energy. Often the corresponding colour of light will be in the UV and there will be no visible colour.

The essential point is that colour comes from electronic transitions that vary a great deal with the atoms involved in the bond. And in many colourful compounds those transitions involve molecular orbitals including multiple atoms across the whole molecule. There is no simple relationship between atoms and the colours of molecules. Every bond is (very) different.


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