Question

Three atoms have following electron configurations respectively:
(i) ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{1}}}$
(ii) ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^5}$
(iii) ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}$
(a) Which of the three has the largest ${{\text{(IE)}}_{\text{1}}}$?
(b) Which has the largest ${{\text{(IE)}}_4}$?
Write your answer as a two digit number.
Ex- if answer for a and b is $3$ and $2$ respectively, answer as $32$ .

Answer 1

Hint: Ionization potential is the energy required for the removal of an electron from the outermost shell of an isolated gaseous atom. The completely filled or half-filled electronic configuration represents the more stable atom.

Complete answer:
Ionization potential is defined as the energy required for the removal of an electron from the outermost shell of an isolated gaseous atom. It is also known as ionization energy.
On going left to right in a period, the ionization potential increases.
The reason for increases in ionization potential is as follows:
1. Increase in effective nuclear charge
2. Decreases in size.
On going down in a group, the ionization potential decreases.
The reason for decreases in ionization potential is as follows:
1. Decrease in effective nuclear charge
2. Increases in size.

a) The largest ${{\text{(IE)}}_{\text{1}}}$ means we have to determine more stable electronic configuration. Because from a more stable configuration it will be hard to remove electrons so the ionization energy will be larger.
In configuration (i),${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{1}}}$, outermost p-orbital has only one electron which can be remove easily because after the removal it will become stable configuration.
In configuration (ii), ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^5}$ , outermost p-orbital has five electrons so, one can be remove easily because it is a unstable configuration.
In configuration (iii),${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}$, outermost s-orbital has one electrons. The fully filled and half-filled electronic configuration for s-orbital is ${{\text{s}}^{\text{2}}}$ and ${{\text{s}}^{\text{1}}}$ respectively. So, it is a stable configuration. It is hard to remove electrons from this configuration so, the (iii) has the largest ${{\text{(IE)}}_{\text{1}}}$.

(b)The largest ${{\text{(IE)}}_4}$ means ionization energy for the fourth electron. We have to determine more stable electronic configuration after the removal of three electrons.
The electronic configuration after the removal of three electrons are as follows:
(i)${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$
(ii)${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^2}$
(iii)${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^4}$
In configuration (i), ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$ , outermost p-orbital has six electrons. The fully filled and half-filled electronic configuration for p-orbital is ${{\text{p}}^{\text{6}}}$ and ${{\text{p}}^{\text{3}}}$ respectively. It is hard to remove electrons from this configuration so, the (i) has the largest ${{\text{(IE)}}_4}$.
In configuration (ii), ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^2}$, outermost p-orbital has two electrons so, one can be remove easily because it is a unstable configuration.
In configuration (iii), ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^4}$, outermost p-orbital has four electrons so, one can be remove easily because it is a unstable configuration. So, the (i) has the largest ${{\text{(IE)}}_4}$.

Therefore the correct answer is $31$.

Note:
The atom having high charge density and small size require more energy to ionize. The atom having low charge density and large size require less energy to ionize. Due to increase in effective nuclear charge and decreases in size, the stabilization of the outermost shell increases, so it becomes hard to remove an electron from the outermost shell on going left to right in a period. Due to decrease in effective nuclear charge and increases in size, the stabilization of the outermost shell decreases, so it becomes easy to remove an electron from the outermost shell on going down in a group. Some elements disobey general trends such as nitrogen or alkaline earth metals due to fully-filled and half-filled electronic configuration.