Question

Element having highest I.P. among the following is:
A.\[{\rm{H}}\]
B.\[{\rm{Li}}\]
C.\[{\rm{B}}\]
D.\[{\rm{Na}}\]

Answer 1

Hint: I.P. represents ionization potential. It is the energy required to remove the last electron from an element. For this we must know the electronic configuration of each element and compare in which of them electrons are most tightly held. That element will have the highest I.P.

Complete step by step answer:
Ionization potential is the energy required to remove an electron from the valence shell of an isolated neutral gaseous atom. It depends on the interaction of the nucleus with the outermost electron, which is also called valence electron. More is the interaction between nucleus and electron; it would be more difficult to remove an electron. Hence higher will be the ionization energy required. Let us look at the configuration of each atom:
Electronic configuration of \[{\rm{H}}\] is \[{\rm{1}}{{\rm{s}}^1}\]
Electronic configuration of \[{\rm{Li}}\] is \[{\rm{1}}{{\rm{s}}^2}2{s^1}\]
Electronic configuration of \[{\rm{B}}\] is \[{\rm{1}}{{\rm{s}}^2}2{s^2}\]
Electronic configuration of \[{\rm{Na}}\] is \[{\rm{1}}{{\rm{s}}^2}2{s^2}2{p^6}3{s^1}\]
\[{\rm{Li}}\] and \[{\rm{Na}}\] belong to the same group i.e. group 1. We know as we move down the group size increases and hence attraction of the nucleus towards the outer electron decreases due to shielding effect. So \[{\rm{Li}}\] has higher ionization energy than\[{\rm{Na}}\]. Option D is eliminated here. Now coming to \[{\rm{B}}\], we know \[{\rm{B}}\] and \[{\rm{Li}}\] belongs to the same period i.e. period 2. The size from left to right along a period decreases due to increase in interaction of nucleus with outer electron. So \[{\rm{B}}\] has higher ionization potential than \[{\rm{Li}}\].
We know hydrogen has only 1 electron in its valence shell, if we remove that one then atom will collapse hence we will not consider this.

So, the highest ionization energy is of option C.

Note:
Since the ionization energy is energy required, that means we have to provide external energy to remove the outer electron. So it always has a positive energy. If you remove one electron then it is termed as ionization energy 1 and if you remove 2 electrons then it is termed as ionization energy 2 and so on.