Arrange the following compounds in increasing order of their dipole moment :

$\mathrm{HBr}, \mathrm{H}_2 \mathrm{~S}, \mathrm{NF}_3$ and $\mathrm{CHCl}_3$

<p>We need to compare the experimental (or well‐established) dipole moments of the given molecules: $\mathrm{HBr}, \mathrm{H_2S}, \mathrm{NF_3},$ and $\mathrm{CHCl_3}$. </p> <hr /> <h3>1. <strong>$\mathbf{NF_3}$</strong></h3> <p><p><strong>Structure</strong>: Trigonal pyramidal (like $\mathrm{NH_3}$), but each bond is more polar toward fluorine. </p></p> <p><p><strong>Net dipole moment</strong>: Quite small, about $0.23\text{–}0.24\,D$ (the bond dipoles partly oppose the lone‐pair contribution).</p></p> <h3>2. <strong>$\mathbf{HBr}$</strong></h3> <p><p><strong>Structure</strong>: Simple diatomic. </p></p> <p><p><strong>Dipole moment</strong>: About $0.78\,D$.</p></p> <h3>3. <strong>$\mathbf{H_2S}$</strong></h3> <p><p><strong>Structure</strong>: 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. </p></p> <p><p><strong>Dipole moment</strong>: About $0.95\,D$.</p></p> <h3>4. <strong>$\mathbf{CHCl_3}$</strong> (chloroform)</h3> <p><p><strong>Structure</strong>: Tetrahedral around carbon with three Cl and one H. </p></p> <p><p><strong>Dipole moment</strong>: About $1.01\,D$.</p></p> <hr /> <h3><strong>Ordering from smallest to largest dipole moment</strong></h3> <p>$ \boxed{\mathrm{NF_3} \;<\; \mathrm{HBr} \;<\; \mathrm{H_2S} \;<\; \mathrm{CHCl_3}.} $</p> <p>Hence, the correct choice is:</p> <p>$ \boxed{\text{Option (C)}\quad NF_3 < HBr < H_2S < CHCl_3.} $</p>