JEE Main 2021 · 26 February · morning · Single correct · +4 / −1
Match List - I with List - II.
Choose the most appropriate answer from the options given below :
| List I Electronic configuration | List II Δ ᵢH in kJ mol⁻¹ | ||
|---|---|---|---|
| (a) | 1s²2s² | (i) | 801 |
| (b) | 1s²2s²2p⁴ | (ii) | 899 |
| (c) | 1s²2s²2p³ | (iii) | 1314 |
| (d) | 1s²2s²2p¹ | (iv) | 1402 |
- (a) (a) → (i), (b) → (iii), (c) → (iv), (d) → (iii)
- (b) (a) → (iv), (b) → (i), (c) → (ii), (d) → (iii)
- (c) (a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)
- (d) (a) → (i), (b) → (iv), (c) → (iii), (d) → (ii)
Answer and solution
Answer (c) (a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)
Solution
Chemistry ma’am · 4 steps
Name the four atoms first, then watch for the two famous dips: B below Be and O below N.
- Identify the atoms. 1s² 2s² is Be, 1s² 2s² 2p¹ is B, 1s² 2s² 2p³ is N and 1s² 2s² 2p⁴ is O. All four are in period 2, so ionisation enthalpy should broadly rise from left to right.
- Be versus B. Be loses an electron from a filled 2s orbital, which is closer to the nucleus and more penetrating. B loses its lone 2p electron, which is shielded by the 2s pair and easier to pull off. So B (801) is lower than Be (899).
- N versus O. N has a half-filled 2p³ shell, an extra-stable arrangement. In O, the fourth 2p electron is paired and repelled by its partner, so it leaves more easily. So O (1314) is lower than N (1402).
Configuration Atom Δᵢ H (kJ mol⁻¹) (a) 1s² 2s² Be 899 (ii) (b) 1s² 2s² 2p⁴ O 1314 (iii) (c) 1s² 2s² 2p³ N 1402 (iv), highest (d) 1s² 2s² 2p¹ B 801 (i), lowest - So. The order is N > O > Be > B, giving (a) → (ii), (b) → (iii), (c) → (iv), (d) → (i). That is option (c).
The idea to keep. Across period 2 ionisation enthalpy rises, with two exceptions: group 13 dips below group 2 (2p above 2s), and group 16 dips below group 15 (paired 2p electron).
Answer from the NTA final answer key. Solution by PrepWiser.