Dipole moment — prediction, measurement, and molecular geometry connection

Question

How do you predict whether a molecule has a net dipole moment? Explain using $\text{CO}_2$ , $\text{H}_2\text{O}$ , $\text{NF}_3$ , and $\text{BF}_3$ as examples. Why does $\text{CO}_2$ have zero dipole moment despite having polar bonds?

Dipole Moment Prediction from Geometry

flowchart TD
    A["Does the molecule have polar bonds?"] -->|No| B["Dipole moment = 0"]
    A -->|Yes| C["Is the molecular geometry symmetric?"]
    C -->|Yes| D["Bond dipoles cancel — net dipole = 0"]
    C -->|No| E["Bond dipoles do NOT cancel — net dipole is non-zero"]
    D --> F["Examples: CO2 (linear), BF3 (trigonal planar), CH4 (tetrahedral), SF6"]
    E --> G["Examples: H2O (bent), NH3 (pyramidal), NF3 (pyramidal), CHCl3"]

Solution — Step by Step

A bond dipole exists whenever two atoms with different electronegativities share a bond. The bond dipole points from the less electronegative atom to the more electronegative one.

A molecular dipole moment ( $\mu$ ) is the vector sum of all bond dipoles in the molecule. Even if individual bonds are polar, the molecular dipole can be zero if the vectors cancel out due to symmetry.

\mu = q \times d

where $q$ is the charge magnitude and $d$ is the distance between charges. Units: Debye (D), where $1 \text{ D} = 3.336 \times 10^{-30} \text{ C} \cdot \text{m}$ .

$\text{CO}_2$ is linear ( $180°$ bond angle). Each C=O bond is polar (O is more electronegative), but the two bond dipoles point in exactly opposite directions:

\leftarrow \text{O=C=O} \rightarrow

They cancel perfectly. Net dipole moment = 0 D. This is why $\text{CO}_2$ is non-polar despite having polar bonds.

$\text{H}_2\text{O}$ is bent ( $104.5°$ bond angle) due to two lone pairs on oxygen. The two O-H bond dipoles point toward oxygen, but because the molecule is not linear, they do NOT cancel:

The resultant vector points from the H side toward the O side. Net dipole moment = 1.85 D — water is highly polar.

$\text{BF}_3$ is trigonal planar (120 degrees, no lone pairs on B). The three B-F bond dipoles are symmetrically arranged in a plane and cancel out. Net dipole = 0 D.

$\text{NF}_3$ is trigonal pyramidal (lone pair on N). The three N-F bond dipoles do NOT lie in the same plane. Their resultant does not cancel. But here is the twist: the lone pair on N has its own dipole (pointing away from the bonds), and in $\text{NF}_3$ , the lone pair dipole opposes the resultant bond dipole (since F pulls electrons away from N). This partial cancellation gives $\text{NF}_3$ a low dipole of 0.23 D.

Compare with $\text{NH}_3$ (1.47 D) where the lone pair dipole reinforces the bond dipoles (both point in the same direction, away from H toward N).

Why This Works

Dipole moment is a vector quantity — direction matters as much as magnitude. The molecular geometry determines the arrangement of bond dipoles, and symmetry is the deciding factor. Symmetric molecules (linear with identical terminal atoms, trigonal planar, tetrahedral, octahedral) have zero net dipole even if individual bonds are polar.

JEE Main frequently asks: “Which of the following has zero dipole moment?” The trick is to check geometry. $\text{CO}_2$ , $\text{BF}_3$ , $\text{CH}_4$ , $\text{CCl}_4$ , $\text{SF}_6$ , $\text{BeF}_2$ — all have zero dipole. $\text{H}_2\text{O}$ , $\text{NH}_3$ , $\text{NF}_3$ , $\text{SO}_2$ , $\text{CHCl}_3$ — all have non-zero dipole.

Alternative Method

Use the VSEPR shortcut: if the molecule has no lone pairs on the central atom AND all surrounding atoms are identical, the dipole moment is zero. If there are lone pairs or the surrounding atoms differ, expect a non-zero dipole.

Molecule	Geometry	Lone Pairs on Central	Identical Ligands?	Dipole?
$\text{CO}_2$	Linear	0	Yes	Zero
$\text{H}_2\text{O}$	Bent	2	Yes	Non-zero
$\text{BF}_3$	Trigonal planar	0	Yes	Zero
$\text{NF}_3$	Trigonal pyramidal	1	Yes	Non-zero
$\text{CHCl}_3$	Tetrahedral	0	No	Non-zero

Common Mistake

Students memorise that ” $\text{NF}_3$ has a lower dipole moment than $\text{NH}_3$ ” but cannot explain why. The reason is the direction of lone pair dipole relative to bond dipoles. In $\text{NH}_3$ , the lone pair and bond dipoles reinforce each other (both point from H toward N). In $\text{NF}_3$ , the lone pair dipole (pointing away from N) opposes the bond dipoles (pointing from N toward F). This opposition reduces the net dipole. JEE Advanced has asked this exact reasoning.