NEETChemistryChemical Bonding and Molecular Structure
Match List-I with List-II regarding the geometry and number of lone pairs on the central atom. List-I (Species) List-II (Geometry, Lone pairs) (A) XeO ₃ (I) Square planar, 2 (B) IF ₅ (II) See-saw, 1 (C) ICl ₄^- (III) Trigonal pyramidal, 1 (D) TeF ₄ (IV) Square pyramidal, 1 Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
- D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct answer
B. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Step-by-step solution
To determine the geometry and number of lone pairs, we calculate the steric number for each central atom: For XeO ₃ : Xe has 8 valence electrons. Oxygen is divalent, so it is not added. Steric number = 1 2 (8 + 0) = 4 ( sp^3 hybridization). It has 3 bond pairs and 4 - 3 = 1 lone pair. The geometry is trigonal pyramidal. So, (A) matches with (III). For IF ₅ : I has 7 valence electrons. There are 5 monovalent F atoms. Steric number = 1 2 (7 + 5) = 6 ( sp^3d^2 hybridization). It has 5 bond pairs and 6 - 5 = 1 lone pai