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

Which of the following species does not exist?
A. BF3B{{F}_{3}}
B. NF3N{{F}_{3}}
C. PF5P{{F}_{5}}
D. NF5N{{F}_{5}}

Explanation

Solution

The unstable nature of trihalides of nitrogen is due to low polarity of NXN-Xbond and a large difference in the size of NN and XX atoms. Phosphorus can extend its covalency beyond 3 because of the presence of empty 3d orbitals.

Complete step by step solution:
Option A: Truly BF3B{{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 BF3B{{F}_{3}} as Fluorine has electrons to give and Boron has void orbital accessible and the back bonding is of the sort 2Pπ2Pπ2P\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.
N1s2,2s2,2p3N-1{{s}^{2}},2{{s}^{2}},2{{p}^{3}}
It contains 3 electrons in 2p2porbital, because there is no vacant 2d2d orbital (it is not possible as it does not exist) to fill its octet, that means to expand so NN can only form NF3N{{F}^{3}} and not NF5N{{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 NF3N{{F}_{3}} and PF5P{{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 PF5P{{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. NF5N{{F}_{5}}.

Note: BF3B{{F}_{3}} is deficient but since of bigger size of 'F' atom, it can't go through dimerization, so it exists as BF3B{{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.