Arrange the following compounds in increasing order of their dipole moment :
HBr, H₂ S, NF₃ and CHCl₃

Solution

Chemistry ma’am · 4 steps

NF₃ is the surprise here: its lone pair works against the N–F bonds, so it comes last.

1. NF3 nearly cancels. In NF₃ each N–F bond moment points away from N, towards the very electronegative F. The lone pair on N points the other way, so the two effects largely cancel and μ is only about 0.23 D.
2. HBr versus H2S. HBr has one polar bond, with an electronegativity difference of about 0.8. H₂S has weaker S–H bonds, but it is bent, so its two bond moments and its lone pairs add up. That puts H₂S slightly above HBr.
3. CHCl3 adds up. In CHCl₃ the three C–Cl bond moments all lean the same way along the C–H axis, and the H on the opposite side has nothing to cancel them. So it has the largest dipole moment of the four.

   Molecule	Shape	What happens to the bond moments	μ (D)
   NF₃	pyramidal	lone pair opposes the N–F moments	0.23
   HBr	linear	one bond, moderately polar	0.79
   H₂S	bent	two S–H moments and lone pairs add	0.95
   CHCl₃	tetrahedral	three C–Cl moments add, nothing cancels them	1.04

4. So. The increasing order is NF₃ < HBr < H₂S < CHCl₃, which is option (c).

The idea to keep. Dipole moment is a vector sum. Check the shape and the direction of every bond moment and lone pair before you trust electronegativity alone.

Answer from the NTA final answer key. Solution by PrepWiser.