Question

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

Answer 1

The existence can be explained by the size and number of valence electrons an atom has.
Valency: It is defined as the capacity of an atom to react and combine with a particular number of atoms of other elements.

Complete step by step answer:
First of all we will talk about halogens.
Halogens: Elements of group $17$ is known as halogens. Examples of halogens are fluorine, chlorine bromine, iodine, astatine. They show the valency of $- 1$ in most of their compounds.
Here we are given with molecules $N{F_3}$ and $N{F_5}$ .
Valency: It is defined as the capacity of an atom to react and combine with a particular number of atoms of other elements.
Now in these molecules we have two atoms that are nitrogen and fluorine. Nitrogen is an element having atomic number $7$ so its electronic configuration is as $1{s^2},2{s^2}2{p^3}$ means it has no vacant orbitals to show valency greater than $3$ . In molecule $N{F_3}$ the valency of fluorine is $- 1$ and let us take the valency of nitrogen as $x$ and as the molecule is neutral (having no charge) so the balance will be like $x - 3 = 0$ . So $x = + 3$ . Here the valency of nitrogen comes out as three so this molecule can exist. But in the molecule $N{F_5}$ the valency of fluorine is $- 1$ and let us take the valency of nitrogen as $x$ and as the molecule is neutral (having no charge) so the balance will be like $x - 5 = 0$ . So $x = + 5$ . But due to absence of vacant orbitals nitrogen cannot show valency greater than three.
Hence $N{F_3}$ is possible but $N{F_5}$ is not.

Note:
For a given orbital maximum number of electrons it can hold is determined as $2(2l + 1)$ where $l$ is azimuthal quantum number. For $s$ the value of $l$ is zero, for $p$ the value of $l$ is one and so on. So the maximum number of electrons in $s$ orbitals is two.