Among H₂S$\text{H}_2\text{S}$, H₂O$\text{H}_2\text{O}$, NF₃$\text{NF}_3$, NH₃$\text{NH}_3$ and CHCl₃$\text{CHCl}_3$, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is:

Related Formula

Core Logic

Step 1: Compare dipole moments:

• H₂O = 1.85 D$\text{H}_2\text{O} = 1.85\text{ D}$
• NH₃ = 1.47 D$\text{NH}_3 = 1.47\text{ D}$
• CHCl₃ = 1.04 D$\text{CHCl}_3 = 1.04\text{ D}$
• H₂S = 0.95 D$\text{H}_2\text{S} = 0.95\text{ D}$
• NF₃ = 0.23 D$\text{NF}_3 = 0.23\text{ D}$ (lowest dipole moment)

Step 2: Identify molecule X = NF₃$X = \text{NF}_3$.

Step 3: Central atom is N (2s² 2p³$2s^2 2p^3$). It forms 3 single bonds with F atoms and retains 1 lone pair of electrons.

Pattern Recognition

Sees: Comparison between NH₃$\text{NH}_3$ and NF₃$\text{NF}_3$ dipoles. Shortcut: In NF₃$\text{NF}_3$, fluorine's high electronegativity pulls electron density away from the lone pair direction, resulting in an exceptionally low dipole moment (0.23 D).

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure