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

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

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

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

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

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Classification of Elements and Periodicity

Outermost electronic configurations of four elements

A, B, C, D

are given below:
(A)

3 s^{2}

(B)

3 s^{2} 3 p^{1}

(C)

3 s^{2} 3 p^{3}

(D)

3 s^{2} 3 p^{4}

The correct order of first ionization enthalpy for them is:

Answer: (B)

The elements are:

Ionization enthalpy trends:

P (C) has the highest ionization enthalpy due to half-filled stability of $p^{3}$ configuration.
S (D) comes next.
Mg (A) has higher ionization enthalpy than Al (B) because s-electrons have better penetration than p-electrons and are held more tightly.
Al (B) has the lowest ionization enthalpy.

Correct order: (B) < (A) < (D) < (C).

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option