Question:
Match List-I with List-II.
Match List-I with List-II.
Updated On: Apr 13, 2026
- A-II, B-III, C-IV, D-I
- A-IV, B-III, C-II, D-I
- A-III, B-II, C-IV, D-I
- A-III, B-II, C-I, D-IV
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Ionization energy generally increases across a period (left to right) but shows anomalies at half-filled (\(p^3\)) and fully-filled (\(s^2\)) subshells due to extra stability.
Step 2: Detailed Explanation:
For \(n=2\) (Period 2 elements): 1. D (ns²np⁶): Neon (Noble gas). Highest IE due to stable octet \(\to\) 2080 (I). 2. C (ns²np³): Nitrogen. High IE due to half-filled p-orbital \(\to\) 1402 (IV). 3. A (ns²): Beryllium. Higher than Boron due to fully-filled s-orbital penetration \(\to\) 899 (II). 4. B (ns²np¹): Boron. Lower than Beryllium \(\to\) 800 (III).
Step 3: Final Answer:
A-II, B-III, C-IV, D-I.
Ionization energy generally increases across a period (left to right) but shows anomalies at half-filled (\(p^3\)) and fully-filled (\(s^2\)) subshells due to extra stability.
Step 2: Detailed Explanation:
For \(n=2\) (Period 2 elements): 1. D (ns²np⁶): Neon (Noble gas). Highest IE due to stable octet \(\to\) 2080 (I). 2. C (ns²np³): Nitrogen. High IE due to half-filled p-orbital \(\to\) 1402 (IV). 3. A (ns²): Beryllium. Higher than Boron due to fully-filled s-orbital penetration \(\to\) 899 (II). 4. B (ns²np¹): Boron. Lower than Beryllium \(\to\) 800 (III).
Step 3: Final Answer:
A-II, B-III, C-IV, D-I.
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