Question

Which of the following has been arranged in order of decreasing dipole moment?

• A. CH$_3$CI > CH$_3$F > CH$_3$Br > CH$_3$l
• B. CH$_3$F > CH$_3$CI > CH$_3$Br > CH$_3$l
• C. CH$_3$CI > CH$_3$Br > CH$_3$l > CH$_3$F
• D. CH$_3$F > CH$_3$CI > CH$_3$l > CH$_3$Br

Answer 1

Answer: (A)

CH$_3$CI > CH$_3$F > CH$_3$Br > CH$_3$l

Answer 2

Dipole moment ($\mu$) is a product of charge separation ($q$) and bond length ($r$), i.e., $\mu = q \times r$. For methyl halides (CH$_3$X), as we move from F to I, the electronegativity difference ($\Delta EN$) between C and X decreases, which tends to decrease $q$. However, the bond length ($r$) increases significantly.

• CH$_3$F: C-F bond has the highest electronegativity difference, but it's also the shortest bond.
• CH$_3$Cl: C-Cl bond has a smaller electronegativity difference than C-F, but a significantly longer bond length. This combination results in CH$_3$Cl having a slightly higher dipole moment than CH$_3$F.
• CH$_3$Br: C-Br bond has lower $\Delta EN$ and longer bond length than C-Cl.
• CH$_3$I: C-I bond has the lowest $\Delta EN$ and the longest bond length.

Therefore, the order is CH$_3$Cl > CH$_3$F > CH$_3$Br > CH$_3$I