Why is phosphorus (v) oxides molecular formula P4O10 and not P2O5 [duplicate]

Question

I understand that the name derives from the empirical formula; however, I do not understand why it does not exist as $\ce{P2O5}$. Is there a compound that exists with this as its molecular formula?

Answer 1

There is no clear rule about how to name a compound which is actually existing as a dimer, a trimer... or an octamer, as long as it does not make a difference in a reaction equation, which is usually the case.

It does not matter in an equation if you count sulfur as a single atom entity (S) or as its true form (S8). You will find the same quantity in both cases.

The same stands for P2O5/P4O10: the true form is usually P4O10 (depending on the solvent), but the 2 forms give the same result at the end.

Answer 2

The short answer is probably that in any case of stoichiometry it doesn't matter. Much as SteffX explained it above. But while we have it already. There are cases where I think it has at least some benefit if we use the not shortened formula. This would probably be one of these cases. As you may know, the $\ce{P4O10}$ is formed when white Phosphorus $\ce{P4}$ reacts with air. The reason for this is that in the $\ce{P4}$-tetrahedron we have smaller bond angles (60°) than we would expect from a head-on bond formation of three p-type orbitals (90°). Therefore the orbitals cannot overlap perfectly and there is some considerable bond strain.

My former Professor for Inorganic Chemistry is a famous phosphorus chemist and he explained it in much more details back then. But if it's about angles in $\ce{P-P}$-bonds, triangular shapes are some of the worst things that can happen. And for the $\ce{P4}$-tetrahedron (white phosphorus) we have a polyhedron that consists of triangular faces only.

So if we introduce oxygen into the system it could enter between every $\ce{P-P}$-bond to increase the angle and therefore reduce the strain. If you draw this on some paper you will see that a $\ce{P4O6}$ results (sometimes called $\ce{P2O3}$). And what you should be able to see is that you can still connect all $\ce{P}$-atoms to get back the original tetrahedron. So the overall shape did not change. We call this a 'topotactic oxidation', oxidation, where the original shape remains, we simply add something inbetween. And in the final step, the phosphorus is at $\ce{P^3+}$ now, we can even oxidize it up to it's highest possible oxidation state $\ce{P^5+}$ by adding more oxygen. It will attack the terminal positions, so we get four additional oxygens and a final $\ce{P4O10}$. This is a relationship many people often forget about. So it's quite nice to remind them of the original $\ce{P4}$-tetrahedron by not shortening the formula down to $\ce{P2O5}$. And as many suggested above it's just the unit you will see.