# Can the compounds NOF3 and POF3 both exist?Why?

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?

• Think of $\ce{HNO3}$; it should be pretty much the same. Nov 14 '15 at 6:39
• @IvanNeretin is it because of a coordinate covalent bond?
– user14857
Nov 14 '15 at 6:41
• I am not getting.Moreover say NCl5 does not exist.
– user14857
Nov 14 '15 at 6:47
• Well, yes, you may think of it as a coordinate bond $\ce{N-O}$, or a structure with charges like $\ce{N+-O-}$, which is the same. It's been discussed before, maybe a dozen times. chemistry.stackexchange.com/questions/7174/… Nov 14 '15 at 7:04
• $\ce{P}$ cannot extend its octet either; the d-orbitals are too far removed in energy. In fact, the creation process of both from atoms is identical.
– Jan
Nov 14 '15 at 13:26

$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$
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.