Question

Which of the following compounds does not exist?
\[
  A.\;N{F_5} \\
  B.\;P{F_5} \\
  C.\;As{F_5} \\
  D.\;Sb{F_5} \\
 \]

Answer 1

Hint: In the options mentioned above, all the compounds contain fluorine and all the other elements belong to group 15 i.e. Nitrogen family. All the elements belonging to group 15 can form trihalides and pentahalides of the type \[M{X_3}\] and\[M{X_5}\] except one element. Let us find out.

Complete answer:
We can see a common thing that all the compounds mentioned above are pentahalides which form five bonds with the fluorine. The configuration of fluorine is \[1{s^2}2{s^2}2{p^5}\] which means five electrons are present in the outermost shell. Apart from fluorine, other elements involved in these compounds are group 15 elements. This group includes Nitrogen (N), Phosphorous (P), Arsenic (As), Antimony (Sb) and Bismuth (Bi). For 15 group elements, in order to form pentahalides, they need to have vacant –d orbitals. All the elements of this group contain vacant –d orbitals except Nitrogen. The electronic configuration of nitrogen is \[1{s^2}2{s^2}2{p^3}\] which means it has three electrons to share with other elements and also no –d orbitals are present and hence can form trihalides but cannot form pentahalides. It means nitrogen can form \[N{F_3}\] but it cannot form\[N{F_5}\] By the above explanation, we can conclude that the compound \[N{F_5}\] does not exist.

So, the correct answer is “Option A”.

Note:
The nitrogen halides are unstable and also said to be explosive because there is a large difference between the size of nitrogen and halogen atom and hence cannot remain bounded to nitrogen atom whereas \[N{F_3}\] is said to be stable because there isn’t much difference between their sizes and this can also be attributed to their similar electronegativity .