Question

In which of the following pairs, the ionization energy of the first species is less than that of the second:
A. ${{O}^{-}},{{O}^{2-}}$
B. S, P
C. N, P
D. $B{{e}^{+}}$ , Be

Answer 1

The amount of energy required to remove the valence electrons from a gaseous atom is called ionization energy. The molecules which are going to have fully filled or half-filled electrons in their orbitals are going to show high ionization energy.

Complete answer:
- In the question it is asked to find the atoms which are going to have high ionization energy among the given options.
- We have to find the atoms which are having half-filled or fully filled electrons in their orbitals.
- The atomic number of sulphur is 16.
- Therefore, sulphur is going to have 16 electrons in its electronic configuration.
- The electronic configuration of sulphur is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$
- We can remove the electron from the 3p orbital of the sulphur atom very easily.
- Means sulphur is going to have less first ionization energy.
- Coming to the atomic number of phosphorus, it is 15.
Therefore, phosphorus is going to have 15 electrons in their electronic configuration.
- The electronic configuration of phosphorus is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$ .
- We cannot remove from the 3p orbital of the phosphorus atom, because the phosphorus is going to have a half-filled electronic configuration.
- Therefore, the first ionization energy of the sulphur is less than the ionization energy of the phosphorus atom.

So, the correct option is B.

Note:
There is a requirement of more energy to remove the electrons from the atom which is going to have the half-filled electrons in their valence orbitals. We should know the electronic configuration of the atom to know about the requirement of the energy to remove the electrons.