Question

Why is $ NC{l_5} $ an unlikely structure?

Answer 1

Hint: In order to answer this question, we should know about the electronic configuration of nitrogen. We should know about the presence of the number of electrons and the orbital that are present in the nitrogen.

Complete answer:
Let’s first understand about the electronic configuration of nitrogen. According to the periodic table, the nitrogen is located in the group $ 15 $ elements. It is the first element in the group $ 15 $ elements. It is the p-block element. The atomic number of nitrogen is $ 7 $ .
The electronic configuration of nitrogen (N) is
 $ N \to 1{s^2}2{s^2}2{p^3} $
By looking at the electronic configuration of nitrogen atoms, it is observed that it is devoid of any sort of d- orbitals. Therefore, it cannot extend its octet and hence it cannot accept five electrons from the five chlorine atoms whatever the energy state is.
Therefore, $ N $ cannot form the $ NC{l_5} $ compound.

Additional Information:
In the case of phosphorus, $ P $ is also a group $ 15 $ elements but it can form the compound $ PC{l_5} $ because it can form five bonds by using d- orbitals. Hence, it can expand its octet.

Note:
It is noted that $ N $ cannot form $ NC{l_5} $ compound but it can form $ NC{l_3} $ compound as a stable compound to complete its octet. This is due to the reason that nitrogen does not have vacant d- orbitals and hence cannot promote the paired electron in $ 3s $ . So, the maximum number of unpaired electrons that nitrogen can have in its valence shell is three i.e. the electron in the $ 2p $ subshell. Therefore, $ NC{l_5} $ does not exist but $ NC{l_3} $ exists. Also, nitrogen is very small to accommodate five chlorine atoms around the central nitrogen atoms.