Chemistry - Does O2 have a color in the gas phase

Solution 1:

Background

$\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black. In the liquid phase it has a light blue color. This color is due to light absorption by the ground state triplet according to the following equation $$\ce{2 O2(^3\Sigma_{g}) ->C[{h\nu}]\ 2 O2(^1\Delta_{g})}$$ This absorption requires light from the red region of the spectrum (~630 nm). If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen.

Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller. Hence, the probability of photon absorption in the gas phase is much reduced. To the human eye oxygen gas will appear colorless (at normal and reduced pressures). However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption.

Solution 2:

An Oxygen molecule has an even number of electrons, but is paramagnetic. This means that it contains two unpaired electrons. The grounds state is a triplet (symmetry and term symbol $^3\Sigma _g^-$) and two low lying excited states one at ($^3\Sigma _g^+$) at 13121 cm$^{-1}$ (762 nm) and $^1\Delta g$ at 7882 cm$^{-1}$ (1270 nm) above the ground state. Their absorption is in the near infra red. Transitions to these levels are spin and symmetry forbidden so occur only weakly. There are also vibrational transitions in the $^1\Delta g$ state at at 1064 nm (v= 0-1)and approx 920 nm (v=0-2) but again these are not in the visible part of the spectrum.

In high pressure gas and in the liquid phase, oxygen is coloured blue and this cannot be due to these transitions as they are too long a wavelength.

The reason for the colour of liquid oxygen was first proposed in 1933 as being due to a collision induced transition involving two oxygen molecules that are transiently close to one another, just as 'collision' suggests, i.e. the absorbing species is transiently (O2)$_2$. Provided there is enough energy in the photon, the transition excites both molecules, and the photon energy is split between them. The complex formed in a collision can have quintet, triplet or singlet character. The two main transitions giving rise to the blue colour are from the transitions

$$(^3 \Sigma _g^-)_2 + hv-> ^1\Delta _g(v=0) + ^1\Delta _g(v=1).$$

These transitions are around 633 nm (equivalent to 2*7889 cm$^{-1}$) and 577 nm respectively (the extra energy is because one vibrational quantum is excited). The transitions in (O2)$_2$ are rather intense compared to O$_2$ due to symmetry breaking, for example, destroying the inversion symmetry. (There are also transitions to $^1\Sigma g^+$ in the 350-400 nm region but these are weak and outside the range we can observe directly by eye). It is also worth mentioning that emission can be observed from these bands, the most intense is called ‘Dimol emission’.

Collision induced transitions are well studied between atoms, but rarer in molecules. It may seem strange that a transition could occur at all since a molecule is a stable arrangement of nuclei and electrons and no such organised species is formed in a collision. However, it takes approx 10-50 fs for the ground state to be changed into an identifiable excited state when a photon is absorbed. A collision lasts for far longer than this, some 1000’s of femtoseconds (10$^{13}$ collisions s$^{-1}$) thus as far as the photon is concerned, it does not need a molecule, just a collection of electrons with energy levels of the right symmetry and energy that need exist only for a few tens of femtoseconds.