Identify the option in which the molecules are arranged in the correct order of their dipole moments.

• A

  $\mathrm{CO}_2 < \mathrm{H}_2 \mathrm{O} < \mathrm{H}_2 \mathrm{~S}$
• B

  $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr}$
• Correct answer
  C

  $\mathrm{BF}_3 < \mathrm{NF}_3 < \mathrm{NH}_3$
• D

  $\mathrm{CH}_4 < \mathrm{CHCl}_3 < \mathrm{CCl}_4$

Solution

$\mathrm{BF}_3$ is non-polar due to equivalent bonds arranged in symmetrical trigonal planar geometry. Among $\mathrm{NF}_3$ and $\mathrm{NH}_3, \mathrm{NH}_3$ is more polar as the dipole moments of the three $\mathrm{N}-\mathrm{H}$ bonds get added whereas they cancel out each other for the $\mathrm{N}-\mathrm{F}$ bonds in $\mathrm{NF}_3$. Thus, the order will be :- $$ \mathrm{BF}_3 < \mathrm{NF}_3 < \mathrm{NH}_3 $$

Hence, the correct answer is option: 'C'.