Match List – I with List – II.

List – I                                                                     List – II

Electronic configuration of elements              ΔiH  in  kJ  mol−1

(A) 1s22s2                                                               (i) 801

(B) 1s22s22p4                                                         (ii) 899

(C) 1s22s22p3                                                         (iii) 1314

(D) 1s22s22p1                                                         (iv) 1402

Choose the most appropriate answer from the options given below

Option 1 - <p>(A) <strong>→</strong> (ii), (B) <strong>→</strong> (iii), (C) <strong>→</strong> (iv), (D) <strong>→</strong> (i)&nbsp;&nbsp;&nbsp;</p>
Option 2 - <p>(A) <strong>→</strong> (i), (B) <strong>→</strong> (iii), (C) <strong>→</strong> (iv), (D) <strong>→</strong> (ii)</p>
Option 3 - <p>(A) <strong>→</strong> (i), (B) <strong>→</strong> (iv), (C) <strong>→</strong> (iii), (D) <strong>→</strong> (ii)</p>
Option 4 - <p>(A) <strong>→</strong> (iv), (B) <strong>→</strong> (i), (C) <strong>→</strong> (ii), (D) <strong>→</strong> (iii)</p>
16 Views|Posted a year ago
Asked by Shiksha User
1 Answer
P
a year ago
Correct Option - 1
Detailed Solution:

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)

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Chemistry Classification of Elements and Periodicity in Properties 2025

Chemistry Classification of Elements and Periodicity in Properties 2025

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