Which of the following bonds would have the largest dipole moment?

• A. H–Br
• B. H–I
• C. H–F
• D. H–Cl

Answer: (C)

Verified step by step guidance

1. Understand that the dipole moment (\(\mu\)) of a bond depends on two main factors: the magnitude of the charge separation and the distance between the charges. It is given by the formula \(\mu = q \times r\), where \(q\) is the magnitude of the partial charges and \(r\) is the bond length.
2. Recognize that the magnitude of the partial charges (\(q\)) is related to the difference in electronegativity between the two atoms involved in the bond. The greater the electronegativity difference, the more polar the bond and the larger the partial charges.
3. Compare the electronegativities of the halogens involved: Fluorine (F) has the highest electronegativity, followed by Chlorine (Cl), Bromine (Br), and Iodine (I). Hydrogen (H) has a lower electronegativity than all these halogens.
4. Consider the bond lengths: as you move down the group in the periodic table (F to I), the atomic size increases, so the bond length (\(r\)) increases. However, the increase in bond length does not compensate for the decrease in electronegativity difference.
5. Conclude that the H–F bond has the largest electronegativity difference and a relatively short bond length, resulting in the largest dipole moment among the given options.