Question

Arrange Mg, Al, Si, P and S in the correct order of their ionisation potential.

Answer 1

Hint: Generally there is the trend of increase of ionisation potential across a period. But there are two exceptions in 3^rd period in case of Mg, Al and P And S.
Complete step by step answer:
Ionisation potential defines the energy needed in the removal of the valence electrons from an atom. Across a period, ionisation potential increases. This is because, as the nuclear charge increases, the attractive electrostatic force between the valence electrons and nucleus increases. And this results in requirement of more energy to remove electrons.
The given elements, Magnesium (Mg), Aluminium (Al), Silicon (Si), Phosphorus (P) and Sulphur (S) all belongs to the third period and their valence shell electronic configuration are:
Mg - \[3{s^2}\]
Al - \[3{s^2}3{p^1}\]
Si - \[3{s^2}3{p^2}\]
P- \[3{s^2}3{p^3}\]
S- \[3{s^2}3{p^4}\]
As stated in the hint, two exceptions such as Magnesium, Aluminium and Phosphorus and Sulphur in case of ionisation potential.
Let’s try to understand exceptions one by one.
In Mg and Al, Mg has valence electron in 3s orbital and Al has valence electron in 3p orbital. As we know, 3p electrons possess more energy than 3s, therefore, Al needs less energy than Mg.
In P and S, Phosphorus possesses three electrons in the orbital p and forms the stable complex. But Sulphur has four electrons in the p orbital and it is unstable because of repulsion due to the third electron. So, removal of electrons from S is easier than P. So, the ionisation potential of S is less than P.

Therefore, the correct order of ionisation potential is, Al

Note: It is to be noted that students might get confused and follow the general trend of ionisation potential that is Mg has lowest IP and S is highest IP. But, this is not correct.