Question

The electronic configuration of nitrogen is 2, 5. How many electrons in the outer shell of the nitrogen atom are not involved in the formation of a nitrogen molecule?

Answer 1

Hint: While the formation of the nitrogen molecule, each nitrogen atom is deficient of 3 electrons to complete their octet. So, each atom will share 3 electrons to form the bond and the rest of the electrons will not be involved in bond formation.

Complete step by step solution: The nitrogen belongs to group 15 of the periodic table and its electronic configuration is –
$$\text{N}\to \text{1}{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^3\text{ or }\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^3$$
We can see that the inner 1s – orbital is fully filled, so it will not take place in bonding. The outermost orbital of nitrogen contains a total of 5 electrons $$\left(2{\text{s}}^{2}2{\text{p}}^3\right)$$ . The nitrogen needs 3 more electrons to complete its octet.
Now, when two atoms of nitrogen approach each other, they both equally share three electrons each and form a triple covalent bond to complete their octet as shown below.

The electronic configuration of the system will look like this:

Thus, we can see that out of 5 valence electrons only 3 are involved in the bonding during the formation of a nitrogen molecule. So, the remaining 2 electrons remain intact in each nitrogen atom.
Hence, two electrons in the outer shell of each nitrogen atom are not involved in the formation of a nitrogen molecule.

Note: The nitrogen forms a covalent bond with another nitrogen atom by sharing its 3 valence electrons. It cannot completely accept or donate electrons to form an ionic bond due to the presence of stable half-filled p-orbital in its valence shell.