JEE MainChemistryChemical Bonding and Molecular Structure
Match List I with List II List I List II (A) SF ₄ (I) Expanded octet, 2 lone pairs on central atom (B) ClF ₃ (II) Expanded octet, 1 lone pair on central atom (C) I ₃^- (III) Expanded octet, 3 lone pairs on central atom (D) PCl ₃ (IV) Obeys octet rule, 1 lone pair on central atom Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
- B(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- C(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct answer
B. (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Step-by-step solution
Number of lone pairs (LP) on the central atom = Valence electrons - Shared electrons 2 . (A) SF ₄ : Sulphur has 6 valence electrons. It forms 4 single bonds with F. LP = 6 - 4 2 = 1 . Total electrons around S = 4 2 (bond pairs) + 1 2 (lone pair) = 10 . This is an expanded octet with 1 lone pair. Matches (II). (B) ClF ₃ : Chlorine has 7 valence electrons. It forms 3 single bonds with F. LP = 7 - 3 2 = 2 . Total electrons around Cl = 3 2 + 2 2 = 10 . This is an expanded octet with 2 lone pairs. Matches (I). (C) I ₃^-