Question

$N{{F}_{3}}$ is stable. But $NC{{l}_{3}}$ is explosive. Identify the reason:
(a) $NC{{l}_{3}}$ undergoes hydrolysis whereas $N{{F}_{3}}$ does not.
(b) $NC{{l}_{3}}$ decomposes on heating and $N{{F}_{3}}$ does not.
(c) $NC{{l}_{3}}$ is a gaseous substance and $N{{F}_{3}}$ is liquid
(d) $NC{{l}_{3}}$ undergoes sublimation while $N{{F}_{3}}$ does not.

Answer 1

The bond between the N-F in $N{{F}_{3}}$ is stable due to almost similar sizes of the nitrogen and fluorine atoms and their overlapping is good as they belong to the same period while the bond between N-Cl in $NC{{l}_{3}}$ is unstable due to difference in the sizes of the nitrogen and the chlorine atoms and their overlapping is not so good as they belong to different periods .Now answer it.

Complete Solution :
- Nitrogen is non- metal and lies in the second period and fifteenth group of the periodic table and has the atomic number as 7 and mass number as 14. It has three unpaired electrons in the valence shell and can gain three electrons in order to complete its octet. On the other hand, halogens i.e. F, Cl etc. has 7 electrons in its outermost valence shell and requires only one electron to complete its octet.
- In, $N{{F}_{3}}$ the 2p orbitals of nitrogen which consists of three unpaired electrons forms three N-F bonds with the 2p orbitals of each fluorine atom each consisting of one unpaired electron. On the other hand , in $NC{{l}_{3}}$, the 2p orbitals of nitrogen which consists of three unpaired electrons forms three N-Cl bonds with the 3p orbitals of each chlorine atom each consisting of one unpaired electron.
- In, $N{{F}_{3}}$, both the nitrogen and fluorine have almost similar size and can form stable bond with each other and their orbitals can easily overlap as they belong to the same period i.e. second period of periodic table and hence, the compound $N{{F}_{3}}$ is stable and is not easily decomposed off when exposed to heating.
- On the other hand, in $NC{{l}_{3}}$, nitrogen has larger size than the fluorine atom and hence, bond between them is unstable and their orbitals do not overlap properly as they belong to different periods of periodic table i.e. nitrogen belongs to second period and chlorine belongs to third period and hence the compound $NC{{l}_{3}}$ is unstable and is easily decomposed when exposed to heating and is, thus, explosive.
Hence, $N{{F}_{3}}$ is stable. But $NC{{l}_{3}}$ is explosive due to the easily decomposition of $NC{{l}_{3}}$ on heating as compared to $N{{F}_{3}}$.
So, the correct answer is “Option B”.

Note: Both $NC{{l}_{3}}$and $N{{F}_{3}}$ has $s{{p}^{3}}$ hybridization and $N{{F}_{3}}$ has smaller bond angle than the $NC{{l}_{3}}$ because as we know that F is more electronegativity than the Cl , so the electron cloud shifts towards chlorine and there is lesser Bond pair- bond pair repulsions in $N{{F}_{3}}$ and hence, has lesser bond angle and is ${{102.5}^{\circ }}$. Bond angle in $NC{{l}_{3}}$ is ${{109.28}^{\circ }}$.