Question

Arrange the following compounds in increasing order of their dipole moment:
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)

• A. CCl\(_3\) < NF\(_3\) < HBr < H\(_2\)S
• B. NF\(_3\) < HBr < H\(_2\)S < CCl\(_3\)
• C. H\(_2\)S < HBr < NF\(_3\) < CCl\(_3\)
• D. HBr < H\(_2\)S < NF\(_3\) < CCl\(_3\)

Hint:

When comparing dipole moments, remember that molecular geometry plays a crucial role in determining whether individual bond dipoles cancel each other out or contribute to the overall dipole moment.

Answer 1

The Correct Option is C

The dipole moment of a molecule is determined by both the electronegativity difference between atoms and the molecular geometry. \begin{itemize} \item CCl\(_3\): Chlorine is highly electronegative, but the molecule has a symmetric trigonal planar geometry, which results in a low dipole moment due to cancellation of individual dipoles. \item NF\(_3\): Nitrogen is more electronegative than fluorine, but due to the geometry of NF\(_3\) (a trigonal pyramidal shape), the dipole moment is moderate. \item HBr: Bromine is less electronegative than fluorine or chlorine, but since HBr has a linear geometry, it results in a moderate dipole moment. \item H\(_2\)S: Due to the bent geometry of H\(_2\)S and the significant electronegativity difference between sulfur and hydrogen, H\(_2\)S has the highest dipole moment among the given compounds. \end{itemize} Thus, the increasing order of dipole moments is: \[ \text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3 \]