Question

Which of the following species does not exist?
A. \[B{{F}_{3}}\]
B. \[N{{F}_{3}}\]
C. \[P{{F}_{5}}\]
D. \[N{{F}_{5}}\]

Answer 1

Hint: The unstable nature of trihalides of nitrogen is due to low polarity of \[N-X\]bond and a large difference in the size of \[N\] and \[X\] atoms. Phosphorus can extend its covalency beyond 3 because of the presence of empty 3d orbitals.

Complete step by step solution:
Option A: Truly \[B{{F}_{3}}\] exists. Indeed it's a Lewis acid. Since it doesn't have its octet complete it is called as hypovalent. Likewise there is Back bonding occurring in \[B{{F}_{3}}\] as Fluorine has electrons to give and Boron has void orbital accessible and the back bonding is of the sort \[2P\pi -2P\pi \] because of which there is fractional double bond character and thus it helps in making boron less electrophilic consequently expanding its stability.
Option B:
The valence of nitrogen is 3.It has 5 electrons in its outermost orbital, but take a look at the electronic configuration.
\[N-1{{s}^{2}},2{{s}^{2}},2{{p}^{3}}\]
It contains 3 electrons in \[2p\]orbital, because there is no vacant \[2d\] orbital (it is not possible as it does not exist) to fill its octet, that means to expand so \[N\] can only form \[N{{F}^{3}}\] and not \[N{{F}^{5}}\].
Option C: Nitrogen and phosphorus both have 5 electrons in their outermost shell. So they need 3 electrons to finish their octet. So \[N{{F}_{3}}\] and \[P{{F}_{5}}\] exist and both nitrogen and phosphorus show the co-valency of 3. Yet, Nitrogen doesn't have empty d orbitals however phosphorus has void 3d orbital. So it can acknowledge more electrons and can expand its covalency to 5 to make \[P{{F}_{5}}\].
Option D: Nitrogen doesn't have any 2d orbitals in its valence shell. In this manner, it can't broaden its covalency upto five.

Hence, the correct option is D. \[N{{F}_{5}}\].

Note: \[B{{F}_{3}}\] is deficient but since of bigger size of 'F' atom, it can't go through dimerization, so it exists as \[B{{F}_{3}}\]with a fractional negative charge in boron and halfway positive sure charge on fluorine. For the existence of any compound it must be stable enough.