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Questions on polarity look like memory tests: is SO₂ polar, is XeF₄ polar, why is NF₃ less polar than NH₃? They are not. Every one of them is answered by the same three steps: find the shape, draw the bond dipoles, and add them as vectors. This post sets out those steps, checks them against NCERT's data, and works one calculation with numbers.
What is a dipole moment?
When two atoms with different electronegativities share electrons, the shared pair shifts towards the more electronegative atom. The NCERT Class 11 chapter Chemical Bonding and Molecular Structure defines the dipole moment as charge × distance of separation, μ = Q × r, measured in debye (D), where 1 D = 3.33564 × 10⁻³⁰ C m. NCERT names the unit after Peter Debye, who won the 1936 Nobel Prize in Chemistry "for his contributions to our knowledge of molecular structure through his investigations on dipole moments and on the diffraction of X-rays and electrons in gases".
Dipole moment is a vector. In chemistry it is drawn as a crossed arrow pointing from the positive end to the negative end, so H–F has its arrowhead on fluorine.
Dipole moment is named in the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus alongside VSEPR theory, and the CBSE 2026-27 chemistry curriculum includes the chemical bonding unit in which NCERT teaches it. Check the current year's syllabus for your exam.
Why can a molecule with polar bonds have zero dipole moment?
Because the bond dipoles can cancel. NCERT's examples:
BeF₂ is linear. The two Be–F dipoles point in opposite directions, so the total is zero.
BF₃ is trigonal planar, with B–F bonds at 120°. The resultant of any two bond dipoles is equal and opposite to the third, so the total is zero.
H₂O is bent, with an H–O–H angle of 104.5°. The two O–H dipoles do not cancel, and NCERT gives a net 1.85 D.
NCERT's table of dipole moments shows the pattern clearly: CO₂, BF₃, CH₄ and CCl₄ all have 0 D despite polar bonds, while CHCl₃, with one C–H in place of a C–Cl, has 1.04 D.
How do you predict polarity in three steps?
Find the shape with VSEPR. Count bonding pairs and lone pairs on the central atom. Our post on VSEPR and molecular shapes has the method.
Draw each bond dipole towards the more electronegative atom, including the effect of lone pairs.
Add them as vectors. If the arrangement is symmetric and all the outer atoms are the same, the sum is zero. Otherwise the molecule is polar.
Try it on a few common exam molecules:
SO₂: bent, so polar.
SF₄: see-saw, with one lone pair, so polar.
XeF₄: square planar, with the two lone pairs opposite each other, so nonpolar.
PCl₅: trigonal bipyramidal with five identical bonds, so nonpolar.
cis- and trans-1,2-dichloroethene: in the trans isomer the two C–Cl dipoles point opposite ways and cancel; in the cis isomer they add, so only cis is polar.
Why is NF₃ less polar than NH₃?
This is the classic trap. Fluorine is more electronegative than nitrogen, so you might expect NF₃ to be more polar. NCERT gives NH₃ = 4.90 × 10⁻³⁰ C m (1.47 D) but NF₃ only 0.80 × 10⁻³⁰ C m (0.23 D in its table).
Both are pyramidal with a lone pair on nitrogen. The lone pair has its own "orbital dipole" pointing away from nitrogen. In NH₃, the N–H bond dipoles point towards nitrogen, so their resultant points the same way as the lone-pair dipole and the two add. In NF₃, the N–F dipoles point away from nitrogen towards fluorine, so their resultant points opposite to the lone-pair dipole and they partly cancel.
Worked example: the O–H bond dipole in water
Water's net dipole moment is 1.85 D and its bond angle is 104.5°. Find the dipole moment of one O–H bond.
Set up the vectors. Each O–H bond dipole μ₁ makes an angle of 104.5°/2 = 52.25° with the line that bisects the H–O–H angle.
Resolve. The components perpendicular to the bisector cancel. The components along it add: μ = 2μ₁ cos 52.25°.
Solve. cos 52.25° ≈ 0.612, so μ₁ = 1.85 ÷ (2 × 0.612) ≈ 1.51 D.
Check the units. 1.85 D × 3.33564 × 10⁻³⁰ C m per D = 6.17 × 10⁻³⁰ C m, the value NCERT quotes.
This is the vector addition you use for forces. At 180°, cos 90° = 0 and the net dipole would vanish, as it does for CO₂.
Why do students find this hard?
Chemistry education researchers Furió and Calatayud (1996) analysed what senior school and university students need to know and be able to do about the geometry and polarity of molecules. They argued that the difficulties go beyond isolated misconceptions to the ways of reasoning students use, which is why a fixed procedure helps. Reasoning from "F is very electronegative, so NF₃ must be very polar" skips the vector step entirely.
Which mistakes cost marks?
Judging polarity from bonds alone. CCl₄ has four polar bonds and zero dipole moment.
Ignoring lone pairs. They decide the shape and add their own dipole, as in NH₃ versus NF₃.
Calling every AB₄ molecule nonpolar. SF₄ is AB₄ but see-saw, so it is polar.
Using the wrong angle. Split the bond angle in half before resolving.
Polarity also controls boiling points and solubility; our post on hydrogen bonding and intermolecular forces picks up from here.
When a shape refuses to make sense on paper, the Fermi Concept Library covers every concept in chemistry, each with its own lesson, so you can go back to VSEPR before tackling polarity.
A practice routine
Day 1: Draw shapes and bond dipoles for ten molecules from NCERT's table and predict which are polar.
Day 2: Use the water method in reverse: find the bond angle at which two 1.5 D bond dipoles give a net 1.0 D.
Day 3: Explain NH₃ versus NF₃ in four sentences without notes.
Day 4: A mixed set of 15 polarity questions, writing the shape before each answer.
Today, take five molecules you would call "obviously polar" and test each with the three steps.
