Question

From the electronic configuration of the given element K,L,M and N , which one has highest ionization potential:
A. $K = [Ne]3{s^2}3{p^2}$
B. $L = [Ne]3{s^2}3{p^3}$
C. $M = [Ne]3{s^2}3{p^1}$
D. $N = [Ar]3{d^{10}}4{s^2}4{p^3}$

Answer 1

Hint: Ionization potential also named as ionisation energy. It is the amount of energy that is required to remove a loosely bound electron from the valence shell of an isolated gaseous atom.
Ionization potential is calculated in electron volt ( $eV$ )

Complete step by step answer:
From the given electronic configuration like:
$K = [Ne]3{s^2}3{p^2}$
It has two electrons in its outermost shell, the valence shell. If it will gain one electron so it will become stable because it will be half filled but to lose an electron it will take some more energy and then also loses the second one electron easily to be stable.
Now in case of $L$ ,
$L = [Ne]3{s^2}3{p^3}$
It has three electrons in its outermost shell that is p-orbital. It means it is half filled so if it loses one electron then it will become unstable so it does not remove electrons from its valence shell easily. It will require the highest energy to lose the electron. So the ionisation energy of this element will be highest.
In the case of $M$ ;
$M = [Ne]3{s^2}3{p^1}$
Here, the outermost shell has only one electron that means it is partially filled. It will get more stable when it loses the electron. So to reach the stability it will lose the electron easily and hence have lowest ionisation energy.
Now, in case of $N$ :
$N = [Ar]3{d^{10}}4{s^2}4{p^3}$
This element has the highest atomic number and lies down to the group in the periodic table. As we know on going down the group ionisation energy decreases because it loses electrons easily due to low shielding effect. This element has lowest ionisation energy.
Hence option B is correct.

Note:
Generally, ionization energy increases from left to right along a period, but there are special cases against this general rule. For example, the ionization energy of Boron is in fact lower than the one of Beryllium while the general trend indicates the first ionization energy of Boron is higher than beryllium. The reasons are:
1) In case of beryllium, electron should be removed from stable configuration of ${s^2}$
so it will take higher energy to remove
2) Similarly Nitrogen has half-filled p-orbitals , ${p^3}$ configuration which is stable than that of Oxygen.