
Nitrogen does not form $N{F_5}$ because:
(A) Nitrogen is a member of the V group.
(B) It contains no empty d-orbital.
(C) The bond energy of $N \equiv N$ is very high.
(D) Inert pair effect exists in the molecule.
Answer
526.5k+ views
Hint: In the given question, we need to understand the covalency of the nitrogen i.e. how many bond formations is possible for nitrogen atoms, from its electronic configuration. It is easy to determine the covalency from the orbitals available in the nitrogen atom for the formation of bonds with another atom.
Complete step-by-step answer:
Let’s first start with the writing of the electronic configuration of the nitrogen.
The atomic number of nitrogen is $7$ and has the electronic configuration as $1{s^2}2{s^2}2{p^3}$ . The common oxidation states of nitrogen are $ + 3, + 5$ .
Nitrogen being the first member of the family has smaller size than others, high electronegativity and high ionization enthalpy. It does not show allotropy.
So, from the above configuration it is clear that there are three $2p$ orbitals available for the formation of bonds with another atom. Because a P-orbital has a capacity of 6 electrons in it. As nitrogen atom contains two shells i.e. K and L which have the capacity of an electron to accommodate is 2 and 8 respectively. Hence, nitrogen has no availability of the d-orbital, that’s why it is not possible to form bonds with more than three atoms. So, it is now clear that nitrogen can make bonds with the three atoms only.
Hence, we can say that nitrogen does not form $N{F_5}$, Because of the unavailability of d-orbital.
So,Option B is correct.
Note:
It is worth noting students have to remember the electronic configuration according to how to fill the electrons in orbit by using the Hund’s rule, Aufbau principle and Pauli’s exclusion principle to confirm the covalency of any atom. We must aware about the shells i.e. K, L, M, …. All the trihalides of the group V elements are stable except those of nitrogen. In case of nitrogen, only $N{F_3}$ is most stable. In other elements the pentahalides are stable due to the availability of vacant d-orbitals. Pentahalides are more covalent than trihalides.
Complete step-by-step answer:
Let’s first start with the writing of the electronic configuration of the nitrogen.
The atomic number of nitrogen is $7$ and has the electronic configuration as $1{s^2}2{s^2}2{p^3}$ . The common oxidation states of nitrogen are $ + 3, + 5$ .
Nitrogen being the first member of the family has smaller size than others, high electronegativity and high ionization enthalpy. It does not show allotropy.
So, from the above configuration it is clear that there are three $2p$ orbitals available for the formation of bonds with another atom. Because a P-orbital has a capacity of 6 electrons in it. As nitrogen atom contains two shells i.e. K and L which have the capacity of an electron to accommodate is 2 and 8 respectively. Hence, nitrogen has no availability of the d-orbital, that’s why it is not possible to form bonds with more than three atoms. So, it is now clear that nitrogen can make bonds with the three atoms only.
Hence, we can say that nitrogen does not form $N{F_5}$, Because of the unavailability of d-orbital.
So,Option B is correct.
Note:
It is worth noting students have to remember the electronic configuration according to how to fill the electrons in orbit by using the Hund’s rule, Aufbau principle and Pauli’s exclusion principle to confirm the covalency of any atom. We must aware about the shells i.e. K, L, M, …. All the trihalides of the group V elements are stable except those of nitrogen. In case of nitrogen, only $N{F_3}$ is most stable. In other elements the pentahalides are stable due to the availability of vacant d-orbitals. Pentahalides are more covalent than trihalides.
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