JEE MainChemistryChemical Bonding and Molecular Structure
Match List I with List II. List I (Molecule) List II (Dipole moment) (A) NH ₃ (I) 0 D (B) H ₂ O (II) 0.23 D (C) BF ₃ (III) 1.47 D (D) NF ₃ (IV) 1.85 D Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- B(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- C(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- D(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Correct answer
D. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Step-by-step solution
The net dipole moment of a molecule is the vector sum of its bond dipoles and lone pair dipoles. For BF ₃ , the central boron atom has 3 bond pairs and 0 lone pairs. Its geometry is trigonal planar, which is perfectly symmetrical. The vector sum of the three B-F bond dipoles is zero. Hence, its dipole moment is 0 D . This matches (C) with (I). For NH ₃ and NF ₃ , both have a pyramidal geometry with one lone pair on the nitrogen atom. In NH ₃ , the orbital dipole due to the lone pair is in the same direction as the