JEE Main 2026 — Classification of Elements and Periodicity in Properties Question with Solution

Given $n=2$, the electronic configurations correspond to elements of the second period.

A. $n{s}^2\Rightarrow 2s^2$ is Beryllium (Be).
B. $n{s}^{2}n{p}^1\Rightarrow 2s^{2}2p^1$ is Boron (B).
C. $n{s}^{2}n{p}^3\Rightarrow 2s^{2}2p^3$ is Nitrogen (N).
D. $n{s}^{2}n{p}^6\Rightarrow 2s^{2}2p^6$ is Neon (Ne).

The first ionization energy generally increases across a period from left to right. However, Beryllium has a higher ionization energy than Boron due to the stable fully-filled $2s$ subshell. Nitrogen has a high ionization energy due to its stable half-filled $2p$ subshell. Neon, being a noble gas, has the highest ionization energy.

The order of first ionization energy is B $<$ Be $<$ N $<$ Ne.

Arranging the given values in increasing order: $800<899<1402<2080$ kJ mol${}^{-1}$.

Thus, the matching is:
A. Be $\to 899$ (II)
B. B $\to 800$ (III)
C. N $\to 1402$ (IV)
D. Ne $\to 2080$ (I)

This corresponds to A-II, B-III, C-IV, D-I.

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