NTA Abhyas JEE Main2020ChemistryChemical Bonding and Molecular StructurePractice
Match the column. Column I Coulmn II (1) S F 2 (p) s p 3 and bent (2) X e F 4 (q) Two lone pairs on central atom (3) N O C l (r) Bond angle < 109.5 ° (4) N F 3 (s) s p 2 and bent (t) s p 3 d 2 and square planar
Options
- A(1)-p, q, r; (2)-q, r, t; (3)-s; (4)-r
- B(1)-p, q; (2)-q, s, t; (3)-s; (4)-r
- C(1)-p, q, r; (2)-q, s, t; (3)-r; (4)-s
- D(1)-p, q; (2)-q, r, t; (3)-r; (4)-s
Correct answer
A. (1)-p, q, r; (2)-q, r, t; (3)-s; (4)-r
Step-by-step solution
S F 2 → s p 3 θ < 109.208 (bent) X e F 4 → s p 3 d (Square planar) N O C l → s p 2 (bent) N F 3 → s p 3 (Pyramidal)