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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. \(1402\)
Choose the correct answer from the options given below:

  1. A A-II, B-III, C-IV, D-I
  2. B A-IV, B-III, C-II, D-I
  3. C A-III, B-II, C-IV, D-I
  4. D A-III, B-II, C-I, D-IV
Verified Solution

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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