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. $ns^2$	I. $2080$
B. $ns^2np^1$	II. $899$
C. $ns^2np^3$	III. $800$
D. $ns^2np^6$	IV. $1402$

Choose the correct answer from the options given below:

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

A. $ns^2 \Rightarrow 2s^2$ is Beryllium (Be).
B. $ns^2np^1 \Rightarrow 2s^2 2p^1$ is Boron (B).
C. $ns^2np^3 \Rightarrow 2s^2 2p^3$ is Nitrogen (N).
D. $ns^2np^6 \Rightarrow 2s^2 2p^6$ 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 $\rightarrow 899$ (II)
B. B $\rightarrow 800$ (III)
C. N $\rightarrow 1402$ (IV)
D. Ne $\rightarrow 2080$ (I)

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

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