Outermost electronic configurations of four elements A , B , C , D are given below: (A) 3   s 2 (B) 3 s…

(A) $3{ \mathrm{s}}^2\to \mathrm{Mg}$

(B) $3{ \mathrm{s}}^{2}3{\mathrm{p}}^1\to \mathrm{Al}$

(C) $3{ \mathrm{s}}^{2}3{\mathrm{p}}^3\to \mathrm{P}$

(D) $3{ \mathrm{s}}^{2}3{\mathrm{p}}^4\to \mathrm{S}$

$\mathrm{Al}$ has one unpaired electron in its highest energy orbital $\left(3\mathrm{p}\right)$, and $\mathrm{Mg}\text{'}\mathrm{s}$ highest energy orbital $\left(3\mathrm{s}\right)$ has the paired electrons.

It is energetically favourable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Compared to $16\mathrm{th}$ group, $15$ group have high ionisation energy. It is because of the stable half filled configuration of he $15\mathrm{th}$ group elements.

$\mathrm{P}-3{\mathrm{s}}^{2}3{\mathrm{p}}^3$

$\mathrm{S}-3{\mathrm{s}}^{2}3{\mathrm{p}}^4$

Therefore, P has more ionisation energy than S.

Hence, the correct order of first ionization enthalpy is $\mathrm{B}<\mathrm{A}<\mathrm{D}<\mathrm{C}$