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.

 \(CCl(_3)\): 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.

 \( NF_3\): Nitrogen is more electronegative than fluorine, but due to the geometry of \(NF_3\) (a trigonal pyramidal shape), the dipole moment is moderate. 

 HBr: Bromine is less electronegative than fluorine or chlorine, but since HBr has a linear geometry, it results in a moderate dipole moment. 

\( H_2S\): Due to the bent geometry of \(H_2S\) and the significant electronegativity difference between sulfur and hydrogen, \(H_2S\) has the highest dipole moment among the given compounds. 

Thus, the increasing order of dipole moments is: \[ \text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3 \]