Recall the general trend of first ionisation enthalpy

Across a period, first ionisation enthalpy generally increases from left to right due to increasing effective nuclear charge.

Here the elements are all in Period 3: $\text{Al},\ \text{Mg},\ \text{P},\ \text{S}$

Write their electronic configurations

• $\text{Mg} : [Ne]3s^2$
• $\text{Al} : [Ne]3s^2 3p^1$
• $\text{P} : [Ne]3s^2 3p^3$
• $\text{S} : [Ne]3s^2 3p^4$

Compare Mg and Al

Although ionisation enthalpy usually increases across a period, $\text{Al}$ has lower first ionisation enthalpy than $\text{Mg}$ because the electron removed from Al is a 3p electron, which is higher in energy and easier to remove than the 3s electron of Mg.

Therefore: $\text{Al} < \text{Mg}$

Compare P and S

Normally ionisation enthalpy should increase from P to S, but there is an exception:

• $\text{P} : 3p^3$ is half-filled, which is extra stable.
• $\text{S} : 3p^4$ has one paired electron in a $p$ orbital, so electron-electron repulsion makes removal easier.

Therefore: $\text{S} < \text{P}$

Combine all comparisons

From the above: $\text{Al} < \text{Mg} < \text{S} < \text{P}$

Match with the options

• A: $\text{Mg} < \text{S} < \text{Al} < \text{P}$ ❌
• B: $\text{Mg} < \text{Al} < \text{S} < \text{P}$ ❌
• C: $\text{Al} < \text{Mg} < \text{S} < \text{P}$ ✅
• D: $\text{Mg} < \text{Al} < \text{P} < \text{S}$ ❌

Final answer

The correct order of first ionisation enthalpy is: $\boxed{\text{Al} < \text{Mg} < \text{S} < \text{P}}$

So, the correct option is C.