# Spin-forbidden transition iof bromine

In one of the lectures of my spectroscopy course, the professor claimed that there is a strong absorption band in the UV/Vis spectrum of bromine, associated with a singlet-state to triplet-state transition. He asked us to think about why this spin-forbidden transition occurs.

So far I haven't really come up with a satisfactory answer. Nevertheless, here are two of my ideas:

1. Broadly speaking, spin-forbidden does not necessarily mean that a transition is not observed, but rather that the intensity of the spectral line will be considerably diminished. Hence, a spin-forbidden transition could be observed in a spectrum.

2. An electronic transition is usually associated with a vibrational transition. Through the vibrational excitation the geometry of the molecule is changed such that the spin selection rules are "relaxed".

Am I on the right track with these ideas?

• Spin-orbit coupling is significant with heavy atoms, which relaxes the spin selection rule. – orthocresol Mar 14 '19 at 15:27
• Spin orbit coupling (coupling of electron spin momentum and orbital angular momentum) increases as $Z^4$ so this will be large in Bromine compared to fluorine. This interaction mixes the singlet and triplet states which means that the spin selection rule is 'broken'. Bromine only has one vibrational mode so option (2) does not exist. – porphyrin Mar 15 '19 at 8:01
• Thank you for the responses. @porphyrin what exactly do you mean by singlet and triplet states "mixing"? – ABCCHEM Mar 15 '19 at 12:38
• It means that there is an operator (in this case spin-orbit coupling) that allows some triplet character to be included in (mixed in with) the singlet state wavefunction. The new state is now not just singlet but has some triplet character, this means that the spin selection rule is no longer a strict rule. In fact it is always the case in molecules, but usually small in the absence of heavy atoms. It manifests itself as causing singlet triplet (intersystem) crossing from an electronically excited singlet state to a triplet, for example in aromatic and other molecules such as benzene. – porphyrin Mar 15 '19 at 14:22

The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity.

If the selection rule does not hold exactly, just like the in the case of the spin selection rule for heavy atoms, a forbidden transition can be intense. Some examples: the $$\text{B}^3\Pi_u-\text{X}^1\Sigma^+_g$$ electronic transition in $$\ce{Br2}$$ and $$\ce{I2}$$, that are strong when compared to the same transition in $$\ce{F2}$$ and $$\ce{Cl2}$$; and the strong $$^3P_1-^1S_0$$ transition in $$\ce{Hg}$$, while the analogous one in $$\ce{He}$$ is very weak. As $$L$$ and $$S$$ are not good quantum numbers when there is strong spin-orbit coupling, the selection rules involving them can't be used.

The second answer you came up with makes sense, but not for spin selection rule. Consider the parity selection rule for molecules with an inversion center. As the molecule vibrates (in an asymmetric mode) it can be distorted into a geometry that does not have an inversion center, so the selection rule does not hold. In this case, this vibronic transition will be allowed.