Question

The first ionization potential(eV) of Be and B respectively are:
(A) $8.29, 9.32$
(B) $9.32, 9.32$
(C) $8.29, 8.29$
(D) $9.32, 8.29$

Answer 1

Hint:Ionization potential is the measure of the difficulty or the energy required for removing the electron from the atom or the ion or it is the tendency of the atom itself to surrender an electron. We can also say that it is strength that is the attractive force by which the electron is held at a place. We can predict the atomic ionization energy using Bohr’s model of atom also.

Complete answer:
Ionisation potential is the minimum energy which is required for the electron so that it could come out of the influence of the nucleus. It tends to increase with the increase in atomic number so in the period it increases from left to right but decreases down the group for high energy orbitals. First ionisation potential is defined as the energy which is required for the first electron to get removed from the valence orbital of the atom. Now the electron configuration for Be is \[1{{s}^{2}}2{{s}^{2}}\] and for B is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\].
Now seeing the electronic configuration of Be we observe that it has a stable electronic configuration because the s orbital is fully filled while in B the p orbital is neither half filled nor fully filled. So the first ionisation potential for Be will be more than B because according to the order of attraction the 2s orbital is closer to the nucleus in comparison to 2p so we need more energy to remove the electron from s orbital in comparison to p orbital. So the first ionization potential for Be will be 9.32eV and for B will be 8.29eV.

So the correct option is option (D).

Note:The ionization potential process is an endothermic process. In an endothermic the energy is usually absorbed so that the formation of products takes place. It also gives us the idea about the reactivity of the compounds and determines the strength of the chemical bonds. It is measured in the unit of eV or kJ/mol.