To determine the order of dipole moments for the given molecules, we should consider the electronegativities of the atoms involved and the molecular geometry.
1. Electronegativity: Fluorine (F) is more electronegative than Bromine (Br). This means that a bond involving fluorine will typically have a higher dipole moment compared to one involving bromine because the difference in electronegativity between carbon and fluorine is greater.
2. Molecular Geometry:
- Cis Isomers: In cis-isomers, the substituents are on the same side of the double bond, leading to a situation where the individual bond dipoles add up rather than canceling out, resulting in a higher overall dipole moment.
- Trans Isomers: In trans-isomers, the substituents are on opposite sides of the double bond, causing the individual bond dipoles to cancel out more effectively, leading to a lower overall dipole moment.
Given these considerations, let's analyze each molecule:
- III. cis-CHF=CHF: This molecule will have the highest dipole moment. Fluorine is more electronegative, and with both fluorines on the same side (cis configuration), the dipoles add up, giving the highest dipole moment among the given molecules.
- I. cis-CHBr=CHBr: This molecule will have a significant dipole moment as it is also in a cis configuration, but because bromine is less electronegative than fluorine, it will have a lower dipole moment than III.
- II. trans-CHBr=CHBr: This molecule will have the lowest dipole moment due to the trans configuration, where the dipoles cancel each other out. Thus, the net dipole moment is the lowest.
Therefore, the order of decreasing dipole moment is:
III. cis-CHF=CHF > I. cis-CHBr=CHBr > II. trans-CHBr=CHBr
So, the correct answer is:
III [tex]${data-answer}gt;$[/tex] I [tex]${data-answer}gt;$[/tex] II
