JEE Mains · Chemistry · STD 11 - 3. Classification of elements and periodicity in properties
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. \(1402\) |
- A A-II, B-III, C-IV, D-I
- B A-IV, B-III, C-II, D-I
- C A-III, B-II, C-IV, D-I
- D A-III, B-II, C-I, D-IV
Answer & Solution
Correct Answer
(A) A-II, B-III, C-IV, D-I
Step-by-step Solution
Detailed explanation
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.…
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