We need to compare the experimental (or well‐established) dipole moments of the given molecules: $\mathrm{HBr}, \mathrm{H_2S}, \mathrm{NF_3},$ and $\mathrm{CHCl_3}$.

1. $\mathbf{NF_3}$

Structure: Trigonal pyramidal (like $\mathrm{NH_3}$), but each bond is more polar toward fluorine.

Net dipole moment: Quite small, about $0.23\text{–}0.24\,D$ (the bond dipoles partly oppose the lone‐pair contribution).

2. $\mathbf{HBr}$

Structure: Simple diatomic.

Dipole moment: About $0.78\,D$.

3. $\mathbf{H_2S}$

Structure: Bent (like $\mathrm{H_2O}$) but S is less electronegative than O, and the S–H bond angle is wider than the O–H angle in water.

Dipole moment: About $0.95\,D$.

4. $\mathbf{CHCl_3}$ (chloroform)

Structure: Tetrahedral around carbon with three Cl and one H.

Dipole moment: About $1.01\,D$.

Ordering from smallest to largest dipole moment

$ \boxed{\mathrm{NF_3} \;<\; \mathrm{HBr} \;<\; \mathrm{H_2S} \;<\; \mathrm{CHCl_3}.} $

Hence, the correct choice is:

$ \boxed{\text{Option (C)}\quad NF_3 < HBr < H_2S < CHCl_3.} $