1. Identify the elements from the outermost configurations

Given valence-shell configurations:

• $(A): 3s^2$
• $(B): 3s^2 3p^1$
• $(C): 3s^2 3p^3$
• $(D): 3s^2 3p^4$

These correspond to Period 3 elements:

• $3s^2 \to \text{Mg}$
• $3s^2 3p^1 \to \text{Al}$
• $3s^2 3p^3 \to \text{P}$
• $3s^2 3p^4 \to \text{S}$

2. Recall the trend of first ionization enthalpy across a period

In general, first ionization enthalpy increases from left to right across a period because effective nuclear charge increases.

However, there are two important exceptions here:

• Mg vs Al:

  • Mg has configuration $3s^2$
  • Al has configuration $3s^2 3p^1$
  • The electron removed from Al is a $3p$ electron, which is higher in energy and less tightly held than the $3s$ electron of Mg.
  • Therefore, $I_1(\text{Al}) < I_1(\text{Mg})$

• P vs S:

  • P has $3s^2 3p^3$, a half-filled $p$-subshell, which is especially stable.
  • S has $3s^2 3p^4$, where one $p$ orbital contains a paired electron, causing extra electron-electron repulsion.
  • Hence it is easier to remove one electron from S than from P.
  • Therefore, $I_1(\text{S}) < I_1(\text{P})$

3. Arrange the given elements

Using the above:

$I_1(\text{Al}) < I_1(\text{Mg}) < I_1(\text{S}) < I_1(\text{P})$

In terms of $A, B, C, D$:

• $A = \text{Mg}$
• $B = \text{Al}$
• $C = \text{P}$
• $D = \text{S}$

So,

$(B) < (A) < (D) < (C)$

4. Check options

• Option A: $(A) < (B) < (C) < (D)$ ❌
• Option B: $(B) < (A) < (D) < (C)$ ✅
• Option C: $(B) < (D) < (A) < (C)$ ❌
• Option D: $(B) < (A) < (C) < (D)$ ❌

5. Final Answer

The correct order of first ionization enthalpy is:

$\boxed{(B) < (A) < (D) < (C)}$

So the correct option is B.