Question

Match the following molecules with their dipole moments:

List I	List II
\( \mathrm{H_2O} \)	I. 0
\( \mathrm{BF_3} \)	II. 0.23
\( \mathrm{NH_3} \)	III. 1.47
\( \mathrm{NF_3} \)	IV. 1.85

• A. A--IV, B--I, C--II, D--III
• B. A--IV, B--I, C--III, D--II
• C. A--IV, B--III, C--I, D--II
• D. A--III, B--IV, C--II, D--I

Hint:

Molecular Dipole Moment: Depends on shape and bond polarity. Symmetric molecules can have polar bonds but zero net dipole due to cancellation.

Answer 1

The Correct Option is B

• {H2O} is bent; polar bonds do not cancel → high dipole: $\mu = 1.85$ D → Match IV
• {BF3} is trigonal planar; symmetric → net $\mu = 0$ → Match I
• {NH3} is trigonal pyramidal; lone pair adds to polarity → $\mu = 1.47$ D → Match III
• {NF3} also trigonal pyramidal; but bond dipoles oppose lone pair → $\mu = 0.23$ D → Match II

Correct match: A--IV, B--I, C--III, D--II → Option (2)

Answer 2

Step 1: Understand dipole moment and molecular geometry
Dipole moment depends on the difference in electronegativity between atoms and the geometry of the molecule. Molecules with symmetrical geometry often have zero dipole moment, while asymmetrical molecules have non-zero dipole moments.

Step 2: Analyze each molecule
- \( \mathrm{H_2O} \): Bent shape due to lone pairs on oxygen, polar molecule with a high dipole moment of about 1.85 D.
- \( \mathrm{BF_3} \): Trigonal planar and symmetrical, dipole moments cancel out, so net dipole moment is 0.
- \( \mathrm{NH_3} \): Trigonal pyramidal shape with lone pair on nitrogen, dipole moment approximately 1.47 D.
- \( \mathrm{NF_3} \): Also trigonal pyramidal, but due to electronegativity differences and bond dipoles partially canceling, it has a smaller dipole moment around 0.23 D.

Step 3: Match molecules with dipole moments
- \( \mathrm{H_2O} \) → 1.85 D (IV)
- \( \mathrm{BF_3} \) → 0 (I)
- \( \mathrm{NH_3} \) → 1.47 D (III)
- \( \mathrm{NF_3} \) → 0.23 D (II)

Step 4: Conclusion
The correct matching is:
\( \mathrm{H_2O} \) – IV, \( \mathrm{BF_3} \) – I, \( \mathrm{NH_3} \) – III, \( \mathrm{NF_3} \) – II.