Question

$N{F_3}$ is possible but $N{F_5}$ is not, why?

Answer 1

Hint:Nitrogen contains only \[3\] unpaired electrons so it can only form \[3\] single bonds with a monatomic element and can show the maximum oxidation state of \[ + 3\].

Complete step by step answer:
Nitrogen is an element with atomic number $7$ and atomic mass $14$ .It was first discovered and isolated by Scottish physician Daniel Rutherford in $1772$. Nitrogen is the lightest member of group $15$ of the periodic table. Dinitrogen forms about $78\% $ of earth's atmosphere making it the most abundant uncombined element. Moreover ${\text{N}}{{\text{F}}_{\text{3}}}$ is an inorganic compound commonly known as Nitrogen trifluoride. This nitrogen-fluorine compound is a colorless, nonflammable gas with a slightly musty odor. Nitrogen trifluoride is a strong greenhouse gas.
The electronic configuration of $N$ is $1{s^2}2{s^3}2{p^3}.$ $N{F_3}$ is possible because nitrogen contains \[3\] unpaired electrons. These three electrons form single bond with three fluoride atoms and hence forms $N{F_3}$ since it does not contain any \[3\]d-orbital so it cannot show oxidation more than \[ + 3\] because it cannot excite electrons from s-orbital to d-orbital and hence $N{F_5}$ does not form. Hence $N{F_3}$ is possible and $N{F_5}$ does not know to exist.

Additional information:- Nitrogen is present in all organisms as amino acids, nucleic acid and as adenosine triphosphate the human body contains nitrate by mass. Many industrially important compounds like cyanides, nitric acid, ammonia and argentic nitrates contain nitrogen.

Note:
The hybridization of $N{F_3}$ molecule is $s{p^3}$ and one bond position is occupied by a lone pair of elements. $N{F_3}$ is an artless, non-flammable gas with a musty odor. $N{F_3}$ is an extremely strong green-house gas $N{F_3}$ is used in hydrogen fluoride and deuterium fluoride laser, which are types of chemical laser.