5
$\begingroup$

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?

|improve this question
$\endgroup$
  • $\begingroup$ Think of $\ce{HNO3}$; it should be pretty much the same. $\endgroup$ – Ivan Neretin Nov 14 '15 at 6:39
  • $\begingroup$ @IvanNeretin is it because of a coordinate covalent bond? $\endgroup$ – user14857 Nov 14 '15 at 6:41
  • $\begingroup$ I am not getting.Moreover say NCl5 does not exist. $\endgroup$ – user14857 Nov 14 '15 at 6:47
  • 1
    $\begingroup$ 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/… $\endgroup$ – Ivan Neretin Nov 14 '15 at 7:04
  • 1
    $\begingroup$ $\ce{P}$ 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. $\endgroup$ – Jan Nov 14 '15 at 13:26
3
$\begingroup$

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.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy