JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Chemical Bonding and Molecular Structure

Dipole moment is the vector sum of individual bond dipoles, and geometry decides how much of that sum survives.

In $\mathrm{N}{\mathrm{F}}_3$, each N–F bond is strongly polar (F is highly electronegative), but nitrogen's lone pair points opposite to the resultant bond-dipole direction, so the bond dipoles and the lone-pair contribution largely cancel: $\mu (\mathrm{N}{\mathrm{F}}_3)\approx 0.24\mathrm{D}$, the smallest of the four despite having the most polar bonds.

$\mathrm{HBr}$ is a simple linear diatomic with a moderate electronegativity difference: $\mu (\mathrm{HBr})\approx 0.79\mathrm{D}$.

${\mathrm{H}}_2\mathrm{S}$ is bent, and the two S–H bond dipoles add along the molecular bisector without the canceling lone-pair effect that $\mathrm{N}{\mathrm{F}}_3$ has: $\mu ({\mathrm{H}}_2\mathrm{S})\approx 0.95\mathrm{D}$.

$\mathrm{CHC}{\mathrm{l}}_3$ is tetrahedral; the three C–Cl dipoles reinforce each other along the C–H axis (the weak C–H dipole opposes them only slightly), giving the largest net moment: $\mu (\mathrm{CHC}{\mathrm{l}}_3)\approx 1.04\mathrm{D}$.

Increasing order of dipole moment: $\mathrm{N}{\mathrm{F}}_3<\mathrm{HBr}<{\mathrm{H}}_2\mathrm{S}<\mathrm{CHC}{\mathrm{l}}_3$.