Question

The orbitals are involved in bonding when nitrogen exists as a diatomic molecule.
A. One s and one p
B. One s and two p
C. Three p
D. One s and three p

Answer 1

Atoms like Nitrogen, Oxygen and Fluorine belong to the second period and p block. Their valence shell has 2s and 2p orbitals. The valence shell and its orbitals take part during bonding. The electrons that an atom uses in bonding are called valence electrons.

Complete step by step solution:
Hybridization is a hypothetical concept. It was given by Pauling. Hybridization is a mixing of atomic orbitals which have almost the same energy but can have different shapes of an atom to form new orbitals having the same energy and same shape. These new orbitals are known as hybrid orbitals. Half filled, fully filled or vacant orbitals can participate in hybridization because it is a mixing of orbitals not electrons.

The number of hybrid orbital forms is equal to the number of intermixing orbits. The orbitals of valence shells participate in hybridization. Hybridization is an endothermic process. Hybrid orbitals tend to orient in such a way to minimize the repulsion between hybrid orbitals. s subshell have only one orbital, p subshell have 3 orbitals, d subshell have 5 orbitals and f subshell have 7 orbitals. Maximum number of electrons an orbital can possess is 2.

Electronic configuration of Nitrogen is $1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^3}$. During hybridization, only the second shell of Nitrogen participates in the formation of a triple bond with another Nitrogen atom to exist as ${{\text{N}}_2}$ molecule. One 2s orbital and three 2p orbital hybridize to form four ${\text{s}}{{\text{p}}^3}$ orbital. So in ${{\text{N}}_2}$ molecules, 4 orbitals participate namely one s and three p orbitals of nitrogen atom. Among four ${\text{s}}{{\text{p}}^3}$ orbitals; three ${\text{s}}{{\text{p}}^3}$ orbitals participate in triple bond formation and one ${\text{s}}{{\text{p}}^3}$ will have the lone pair of electrons. Similarly ${{\text{O}}_2}$ and ${{\text{F}}_2}$ have 2s and 2p orbitals and their electrons involved in bonding. But on the other hand, atoms like P, S and Cl will have 3s and 3p orbitals and their electrons involved in bonding.

Thus, the correct option is D.

Note: ${{\text{O}}_2}$ , ${{\text{N}}_2}$ and ${{\text{F}}_2}$ are homonuclear diatomics. The ${{\text{H}}_2}$ molecule is the only molecule which uses electrons in its 1s orbitals to form chemical bonds.