Outermost electronic configurations of four elements A, B, C, D are given below:
(A) $$3s^2$$
(B) $$3s^2 3p^1$$
(C) $$3s^2 3p^3$$
(D) $$3s^2 3p^4$$
The correct order of first ionization enthalpy for them is

Elements of the third period have the principal quantum number $$n = 3$$ for their valence-shell electrons. The four given configurations therefore correspond to

$$A : 3s^{2}\; \Longrightarrow\;$$ magnesium (Mg)
$$B : 3s^{2}\,3p^{1}\; \Longrightarrow\;$$ aluminium (Al)
$$C : 3s^{2}\,3p^{3}\; \Longrightarrow\;$$ phosphorus (P)
$$D : 3s^{2}\,3p^{4}\; \Longrightarrow\;$$ sulphur (S)

In a period, the first ionization enthalpy (I.E.) generally increases from left to right because the nuclear charge increases while the valence shell remains the same. However, two well-known exceptions disturb the smooth rise:

Case 1: The jump from a filled $$s$$ subshell ($$ns^{2}$$) to the first $$p$$ electron ($$ns^{2}np^{1}$$).
Removing an electron from the new, higher-energy $$p$$ subshell needs less energy than removing it from the lower-energy, fully-filled $$s$$ subshell. Hence
$$\text{I.E.}(Al) \lt \text{I.E.}(Mg)$$

Case 2: Half-filled and fully-filled subshell stability.
A half-filled $$p^{3}$$ subshell (as in P) is extra stable. When we move from $$p^{3}$$ to $$p^{4}$$ (S), the added electron must pair up in an already occupied orbital, introducing inter-electronic repulsion and slightly lowering the I.E. Therefore
$$\text{I.E.}(S) \lt \text{I.E.}(P)$$

Combining the normal trend with the two exceptions gives the overall order

$$\text{Al} \lt \text{Mg} \lt \text{S} \lt \text{P}$$

Replacing element symbols with the given letters:

$$B \lt A \lt D \lt C$$

Hence the correct order of first ionization enthalpy is described by Option B.

Answer → Option B \bigl(B < A < D < C\bigr)