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I have noticed that liquid $\ce{O2}$ (I clarify it as $\ce{O2}$ because oxygen exists in several other forms which have different colors in the liquid state than $\ce{O2}$) has a light blue color to it.

liquid oxygen

In the solid state it has a light blue color like it does in the liquid state and then finally a red color if enough pressure is applied. It can also be pink, black, or metallic again depending on the pressure and temperature.

Does it also have a light blue color in the gas phase?

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3 Answers 3

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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.

Answer

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.

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  • $\begingroup$ It is a triplet-singlet transition, therefore it is spin forbidden. The transition can only occur if some other process compensate for the magnetic moment, which is more probable in a condensed phase. $\endgroup$
    – Greg
    Commented Aug 9, 2014 at 16:22
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    $\begingroup$ @Greg As mentioned above, the conversion of ground state, triplet oxygen to singlet oxygen involves the simultaneous absorption of one photon by 2 triplets to produce 2 singlets. Therefor, total spin can be conserved and the process is not spin forbidden, e.g. $\ce{ ^ ^ ~(v) (v) -> ^ (v)~ ^ (v) }$ $\endgroup$
    – ron
    Commented Aug 9, 2014 at 18:22
  • $\begingroup$ The goal of my comment was to add the term "spin-forbiden" as it is something OP may heard about and can connect. I did not correct or argued your description. $\endgroup$
    – Greg
    Commented Aug 9, 2014 at 19:57
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    $\begingroup$ @Greg Thanks for the clarification. I didn't think you were correcting my answer, that is why I avoided using words like "right" and "wrong" in my comment. I just wanted to elaborate your comment a bit further within the context of the original question, so that a reader would understand why, in this particular case, the photoconversion is spin-allowed. $\endgroup$
    – ron
    Commented Aug 9, 2014 at 21:57
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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.

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no, in gas phase it's colorless. someone suggests me to be specific, then let me show you oxygen gas's UV absorption spectrum calculated in Gaussian and visualized in GaussView. charge=0 triplet 1 atm 298K zindo charge=0 triplet 1 atm 298K zindo

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  • $\begingroup$ why is it then that O3 is light blue in the gas phase and dark blue in the liquid phase while O2 is light blue as a liquid and colorless as a gas because I always thought that the scattering of blue light and how it gets darker and darker in the sky is partly due to differences in O2 concentration and O3 concentration with O2 contributing more to the lighter blue and O3 contributing more to the darker blue. $\endgroup$
    – Caters
    Commented Aug 9, 2014 at 6:11
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    $\begingroup$ @caters AFAIK the blue color of sky is attributed to different scattering of light of different color on microparticles of dust in the atmosphere. $\endgroup$
    – permeakra
    Commented Aug 9, 2014 at 8:57
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    $\begingroup$ this high CO2 is also the reason why mars has a red sky during parts of the martian year that aren't dust storms(during which the red color is then caused mainly by iron oxide dust) $\endgroup$
    – Caters
    Commented Aug 9, 2014 at 16:05
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    $\begingroup$ I would consider zindo hardly a method of proof, while uv/vis calculations tend to derive from the experiment by a lot. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 10, 2014 at 7:23
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    $\begingroup$ @Ath the calculation method you choose is like picking a number from the phone book. It happens to be close to the experimentally observed, but that is just by coincidence. $\endgroup$
    – Martin - マーチン ♦
    Commented Jul 29, 2015 at 5:33

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