Match List - I with List - II. List - I Electronic configuration of neutral atom (where n=2) A. ns2 B. ns2np1 C. ns2np3 D. ns2np6 List - II 1st Ionization Energy (kJ mol-1) I. 2080 II. 899 III. 800 IV. 1402 Choose the correct answer from the options given below :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q58 – JEE-MAIN 2026

Match List - I with List - II.

\begin{array}{lcl} List - I & List - II \\ Electronic configuration & {1st Ionization Energy (kJ mol}^{- 1}) \\ of neutral atom (where n=2) \\ 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 \end{array}

Choose the correct answer from the options given below :

For $n = 2$ , match the given electronic configurations to specific elements in the second period.

Step 1: Identify Elements from Configurations✦ Active

For $n = 2$ , the given electronic configurations correspond to the following elements:

\begin{array}{ll} A . n s^{2} & \to Be (1 s^{2} 2 s^{2}) \\ B . n s^{2} n p^{1} & \to B (1 s^{2} 2 s^{2} 2 p^{1}) \\ C . n s^{2} n p^{3} & \to N (1 s^{2} 2 s^{2} 2 p^{3}) \\ D . n s^{2} n p^{6} & \to Ne (1 s^{2} 2 s^{2} 2 p^{6}) \end{array}

Step 2: Recall Ionization Energy Trends○ Expand

The general trend for first ionization energy (IE1) across a period is an increase from left to right. However, there are exceptions:

1. IE1 of Group 13 (B) is less than Group 2 (Be) due to the easier removal of a $p$ -electron and increased shielding.

2. IE1 of Group 16 (O) is less than Group 15 (N) due to the extra stability of the half-filled $p$ -orbital in Group 15 elements.

The approximate IE1 values for these elements are: Be (899 kJ/mol), B (800 kJ/mol), N (1402 kJ/mol), Ne (2080 kJ/mol).

💡 Teacher's Secret Hint

Remember the exceptions to the general trend of ionization energy across a period, especially for Group 2 vs 13 and Group 15 vs 16.

Step 3: Match Configurations to Ionization Energies○ Expand

Based on the identified elements and their ionization energies:

\begin{array}{ll} A . n s^{2} (Be) & \to II. 899 {kJ mol}^{- 1} \\ B . n s^{2} n p^{1} (B) & \to III. 800 {kJ mol}^{- 1} \\ C . n s^{2} n p^{3} (N) & \to IV. 1402 {kJ mol}^{- 1} \\ D . n s^{2} n p^{6} (Ne) & \to I. 2080 {kJ mol}^{- 1} \end{array}

Thus, the correct match is A-II, B-III, C-IV, D-I.

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