JEE Main 2026 Apr 8 Shift 2, Chemistry Q8: Periodicity In Atomic Properties

JEE Main2026Apr 8, Shift 2Chemistry

Match List-I with List-II.

List-I (Electronic configuration of neutral atom, where $n = 2$ )	List-II (1 $^{st}$ Ionization Energy, kJ mol $^{- 1}$ )
A. $n s^{2}$	I. $2080$
B. $n s^{2} n p^{1}$	II. $899$
C. $n s^{2} n p^{3}$	III. $800$
D. $n s^{2} n p^{6}$	IV. $1402$

Choose the correct answer from the options given below :

A
A-II, B-III, C-IV, D-I
B
A-IV, B-III, C-II, D-I
C
A-III, B-II, C-IV, D-I
D
A-III, B-II, C-I, D-IV

Solution

For $n = 2$ the configurations correspond to: A → Be ( $2 s^{2}$ ), B → B ( $2 s^{2} 2 p^{1}$ ), C → N ( $2 s^{2} 2 p^{3}$ ), D → Ne ( $2 s^{2} 2 p^{6}$ ).

Ordering ionization energies.

$B$ has the lowest IE among these because removing its single $2 p$ electron is easier than breaking the filled $2 s^{2}$ shell of Be.

$B e$ ( $2 s^{2}$ ) is next — a filled subshell adds stability but the electron is in the outermost $s$ orbital.

$N$ ( $2 s^{2} 2 p^{3}$ ) has extra stability from a half-filled $2 p$ , so its IE exceeds that of Be.

$N e$ ( $2 s^{2} 2 p^{6}$ ) has the highest IE — a fully-filled noble-gas configuration.

Order: B (800) $<$ Be (899) $<$ N (1402) $<$ Ne (2080).

$∴$ A–II, B–III, C–IV, D–I.

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JEE Main 2026 Apr 8 Shift 2, Chemistry Q8: Periodicity In Atomic Properties

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