JEE MainChemistryChemical Bonding and Molecular Structure
Match List I with List II. List I (Molecule) List II (Lone pairs on central atom, Shape) (A) SF ₄ (I) 2 lone pairs, Bent T-shape (B) BrF ₅ (II) 3 lone pairs, Linear (C) XeF ₂ (III) 1 lone pair, Square Pyramidal (D) ClF ₃ (IV) 1 lone pair, See-saw Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- B(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- C(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- D(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Correct answer
A. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Step-by-step solution
For SF ₄ : Central atom S has 6 valence electrons. It forms 4 single bonds with F . Steric number = 6 + 4 2 = 5 ( sp ^3 d hybridization). With 4 bond pairs, it has 1 lone pair. The shape is see-saw. So, (A) matches (IV). For BrF ₅ : Central atom Br has 7 valence electrons. It forms 5 single bonds with F . Steric number = 7 + 5 2 = 6 ( sp ^3 d ^2 hybridization). With 5 bond pairs, it has 1 lone pair. The shape is square pyramidal. So, (B) matches (III). For XeF ₂ : Central atom Xe has 8 valence electrons. It forms 2