Question

The anomalous behavior of Nitrogen is due to:
A.Small size and high electronegativity
B.Non availability of d orbitals in valence shell
C.Ease of multiple bond formation
D.All are correct

Answer 1

Hint: Group 15 elements known as the Nitrogen family. It includes the elements Nitrogen, Phosphorous, Arsenic, Antimony, Bismuth. Among these Nitrogen and Phosphorus are non-metals, Arsenic and Antimony are metalloids, and Bismuth is a typical metal.

Complete step by step answer:
Nitrogen differs from the rest of the elements in the group. This is due to its properties. We know, on moving down the group size of an atom decreases. Thus, Among Group 15 elements Nitrogen is smaller in size.
Nitrogen has very high ionization energy because of its stable half-filled configuration. It has an electronic configuration of \[1{s^2} 2{s^2} 2{p^3}\]. Here P orbital is half-filled. So, it is difficult to remove an electron from this stable configuration.
Nitrogen has high electronegativity value. Generally, electronegativity decreases down the group with increase in atomic size.

Element	N	P	As	Sb	Bi
Electronegativity	3	2.1	2	1.9	1.9

In nitrogen only S and P orbitals are present. Its configuration is \[1{s^2} 2{s^2} 2{p^3}\]. Non-availability of d orbital is also a reason for the anomalous behavior of Nitrogen atoms.
Nitrogen shows the unique property of form \[p\pi - p\pi \] multiple bonds with itself and with other elements. Other members can’t form this bond because their atomic orbitals are large so they cannot have effective overlapping. Nitrogen exists as a diatomic molecule and forms a triple bond between them.
Nitrogen is less reactive. Its Bond order is 3. So, the bond enthalpy of nitrogen is high and it is very stable.
Nitrogen does not have d orbitals in its valence shell. So, they can’t form a dπ-dπ bond. While other members of the group form dπ-dπ bonds with transition metals.
Nitrogen shows a variety of oxidation states due to its small size, high electronegativity, and tendency of forming pπ-pπ bonds. It shows oxidation state ranging from -3 to +5.
The hydride of Nitrogen \[N{H_3}\] is thermally stable and non-poisonous. But hydrides of other members are less stable.

Hence, the correct answer is option (D) i.e All of the above.

Note: Among the trihalides of nitrogen only \[N{F_3}\] is stable. \[NC{l_3}\], \[NB{r_3}\] , NI3 are unstable and reactive. Trihalides of other members are less reactive. Also, Nitrogen does not show pentavalency. It can’t form \[NC{l_5}\] due to the non-availability of d orbitals.