Question

Explain why the first I.E. of Al is lower than that of Mg.

Answer 1

Electronic configuration is:

$Al⟹\left[Ne\right]3s^{2}3p^{1}Mg⟹\left[Ne\right]3s^2$

Therefore, as a single electron is present in outer p-orbital, it can be removed easily. But in $Mg$, the s-orbital is full and high amount of energy as compared to that of $I.E{.}_1$ of $Al$ is needed.

So, the first $I.E{.}_1$ of $Al$ is lower than that of $Mg$.