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Today I came across a question (in one of my books) like
Can the compounds $\ce{NOF3}$ and $\ce{POF3}$ both exist? Why?
After some googling I found that both of them do indeed exist. I can understand that $\ce{P}$ can extend its octet due to empty d orbitals. But $\ce{N}$ can't do so. So how can it possibly exist?
Both molecules can be considered the acid (hydrate) trifluoride of the respective acid (phosphoric acid or nitric acid). In the case of phosphoric acid, this is immediately obvious: Just replace the three hydroxide groups with a fluorine atom, each. For the nitrogen compound, you initially need to (formally) add water before you can do the substitution:
$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$
(Note: this is not a proper chemical reaction, merely an illustration of the thought processes.)
You could also approach this compound from the ammonium cation $\ce{NH4+}$ by replacing three hydrogen atoms with fluorine and the fourth with oxygen. Or you start off with $\ce{NF3}$ (isoelectronic to $\ce{PF3}$ and then oxidise the nitrogen by adding a single oxygen atom much in the same way you would oxidise from $\ce{HNO2}$ to $\ce{HNO3}$.
In fact, for many of the compounds traditionally explained with d-orbital participation for main-group elements, an explanation adhering to the octet rule (no more than eight valence electrons per main-group atom; two for hydrogen/helium) is often closer to the truth. This includes, but is not limited to $\ce{OPF3, SO4^2-}$, sulphonic acids and many more.
