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

Molecular dipole moments are influenced by atomic electronegativity differences and molecular structure.

\(CCl(_3)\): Despite chlorine's high electronegativity, the molecule's symmetric trigonal planar arrangement causes individual dipoles to cancel, yielding a low net dipole moment.

 \( NF_3\): Although nitrogen is more electronegative than fluorine, the trigonal pyramidal geometry of \(NF_3\) results in a moderate dipole moment. 

 HBr: Bromine exhibits lower electronegativity than fluorine or chlorine. However, HBr's linear structure leads to a moderate dipole moment. 

\( H_2S\): The bent geometry of \(H_2S\) and the substantial electronegativity difference between sulfur and hydrogen contribute to it possessing the largest dipole moment among the compounds discussed. 

Therefore, the dipole moments increase in the following order: \[ \text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3 \]