# Why is HBr colorless while Br2 isn't?

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.

• The neutral Br atom has no electronic transitions in the visible range. The Br${2}$ molecule has them, because of vibrational states. Dec 9, 2020 at 19:15
• 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. Dec 9, 2020 at 19:16
• @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. Dec 9, 2020 at 20:10
• Why was this question closed? It is clearly not a homework question and I did share my thoughts Dec 9, 2020 at 21:35
• 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 $\ce{Br2}$. Dec 9, 2020 at 22:12