JEE Main 2026 — Chemistry Inorganic Chemistry

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

A. $ns^2 \Rightarrow 2s^2$ns2⇒2s2 is Beryllium (Be).

B. $ns^2np^1 \Rightarrow 2s^2 2p^1$ns2np1⇒2s22p1 is Boron (B).

C. $ns^2np^3 \Rightarrow 2s^2 2p^3$ns2np3⇒2s22p3 is Nitrogen (N).

D. $ns^2np^6 \Rightarrow 2s^2 2p^6$ns2np6⇒2s22p6 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$2s subshell. Nitrogen has a high ionization energy due to its stable half-filled $2p$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$800<899<1402<2080 kJ mol$^{-1}$−1.

Thus, the matching is:

A. Be $\rightarrow 899$→899 (II)

B. B $\rightarrow 800$→800 (III)

C. N $\rightarrow 1402$→1402 (IV)

D. Ne $\rightarrow 2080$→2080 (I)

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

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