Match First Ionization Energy with Neutral Atom Electronic Configurations

To find the correct match between the electronic configurations of neutral atoms (where $n=2$) and their first ionization energies ($\Delta_i H_1$), we identify the corresponding period-2 elements and analyze their electronic structures:

1. Configuration A: $ns^2$ ($2s^2$, Beryllium - $\text{Be}$)

   • Beryllium has a completely filled $2s$ subshell, which provides extra stability and a penetration effect of $s$-electrons.
   • Its first ionization energy is higher than that of Boron ($2s^2 2p^1$).
   • Therefore, $\text{A} \rightarrow \text{II } (899 \text{ kJ mol}^{-1})$.

2. Configuration B: $ns^2 np^1$ ($2s^2 2p^1$, Boron - $\text{B}$)

   • The electron to be removed lies in a $2p$ orbital, which is shielded by the inner $2s$ electrons and experiences less nuclear attraction than the $2s$ electrons in Be.
   • Thus, its ionization energy is lower than that of Be.
   • Therefore, $\text{B} \rightarrow \text{III } (800 \text{ kJ mol}^{-1})$.

3. Configuration C: $ns^2 np^3$ ($2s^2 2p^3$, Nitrogen - $\text{N}$)

   • Nitrogen has a stable, half-filled $2p$ subshell ($2p_x^1 2p_y^1 2p_z^1$), which requires a higher amount of energy to remove an electron compared to typical non-metals in the same period.
   • Therefore, $\text{C} \rightarrow \text{IV } (1402 \text{ kJ mol}^{-1})$.

4. Configuration D: $ns^2 np^6$ ($2s^2 2p^6$, Neon - $\text{Ne}$)

   • Neon is a noble gas with a completely filled valence octet ($2s^2 2p^6$), conferring maximum stability and extremely high effective nuclear charge.
   • It possesses the highest first ionization energy among the period-2 elements given.
   • Therefore, $\text{D} \rightarrow \text{I } (2080 \text{ kJ mol}^{-1})$.

Summarizing the matches:

• A $\rightarrow$ II
• B $\rightarrow$ III
• C $\rightarrow$ IV
• D $\rightarrow$ I

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