What is the state of the oxygen free radical in the following reaction? $$\ce{O2(g) -&gt;[h\nu][\nu\,&lt;\,240~nm]2O^{.}(?)}$$

Oxygen is split by UV light below 240 nm to form oxygen radicals, which then go on to react with oxygen molecules to form ozone.

Firstly, is the oxygen a free radical?

Oxygen has six outer shell electrons, but my teacher mentioned that by the way orbitals work, they don't actually all pair off, rather you end up with two pairs of electrons and two unpaired electrons, hence making oxygen a free radical, though I'm not too sure. But because the free radical is so highly reactive, I'm not sure whether I'm even allowed to assign it a state, because straight after being formed it pretty much immediately reacts.

Going by process of elimination, it can't be aqueous, because it isn't dissolved in water, and I don't think it is solid, because solids are structurally rigid. I am therefore inclined to say that it is a gas, but my main problem is that each reaction only creates two oxygen free radicals, which immediately disappear, so it shouldn't actually show any properties of being a gas.

So what is the state I should assign the oxygen, or does it simply not need one?
According to the reaction $$\ce{O2\,(g) -&gt;[h\nu][\nu\,&lt;\,240~\mathrm{nm}] 2O}$$
photolytic cleavage of oxygen yields two radicals. These will most likely be in the high spin triplet state. And yes these are free radicals according to the definition of the IUPAC Goldbook, as they have two unpaired electrons.

I assume you are talking about state of matter and then of course, you are right. Oxygen radicals can exist neither in aqueous solution, nor liquid, nor solid. Even in gas phase the concentration of oxygen radicals will be very low as they are an incredibly short lived species. They might exist in plasma, too. Keeping that in mind it is probably unnecessary to assign a state to the oxygen radical.