Arrange the following compounds in increasing order of their dipole moment:
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)
Show Hint
- CCl\(_3\)<NF\(_3\)<HBr<H\(_2\)S
- NF\(_3\)<HBr<H\(_2\)S<CCl\(_3\)
- H\(_2\)S<HBr<NF\(_3\)<CCl\(_3\)
- HBr<H\(_2\)S<NF\(_3\)<CCl\(_3\)
The Correct Option is C
Solution and Explanation
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: \[ \text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3 \]
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