JEE Mains Chemistry · Chemical Bonding and Molecular Structure

Dipole Moment, Hydrogen Bonding and Intermolecular Forces

A molecule's dipole moment is the vector sum of its bond dipoles and lone-pair moments, so symmetry can cancel it; hydrogen bonds and other intermolecular forces then decide how molecules hold together.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Twenty-seven PYQs, nineteen of them multiple choice, and two from 2026. Seven ask about the size or direction of a dipole moment, most often NH₃ against NF₃; nine count or pick polar and non-polar molecules; eleven test hydrogen bonding and other intermolecular forces. Three ideas cover the page.

Concept 1 of 3: Dipole moment: size and direction

When two bonded atoms differ in electronegativity, the shared electrons shift towards one of them, leaving a small positive and a small negative end. The dipole moment is that charge times the distance between the ends. In a molecule, the bond moments and any lone-pair moment add as vectors.

Definition

$\mu = q \times d$ . Unit: debye, $1\ \text{D} = 10^{-18}$ esu cm $= 3.336 \times 10^{-30}$ C m. The electron's charge is $4.8 \times 10^{-10}$ esu.
Chemists draw a crossed arrow from the positive end to the negative end: the cross at the positive end, the head at the negative end, showing where the electron density moves.
$\mathrm{NH_3}$ (1.47 D) > $\mathrm{NF_3}$ (0.23 D). In $\mathrm{NH_3}$ the N–H bond moments point towards N and add to the lone-pair moment. In $\mathrm{NF_3}$ the N–F bond moments point away from N and partly cancel it.
Values to know: $\mathrm{H_2O}$ 1.85, $\mathrm{NH_3}$ 1.47, $\mathrm{CHCl_3}$ 1.04, $\mathrm{H_2S}$ 0.95, HBr 0.79, $\mathrm{NF_3}$ 0.23 D.
Fraction of ionic character $= \dfrac{\mu_{\text{observed}}}{\mu_{\text{fully ionic}}}$ , with $\mu_{\text{fully ionic}} = e \times d$ .

Dipole moment

\mu = q \times d \qquad 1\ \text{D} = 10^{-18}\ \text{esu cm}

Worked example

HCl has a dipole moment of 1.07 D and a bond length of 1.27 Å. Find the partial charge on each atom in esu, and the fraction of an electron's charge it represents (

e = 4.8 \times 10^{-10}

esu).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q36Moderate

Example 1 · Chemical Bonding and Molecular Structure · Dipole Moment, Hydrogen Bonding and Intermolecular Forces

Among

H_{2}\text{ }S,H_{2}O,NF_{3},NH_{3}

and

CHCl_{3}

, identify the molecule

(X)

with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule

(X)

is :

More electronegative F does not mean a bigger dipole

Each N–F bond is more polar than an N–H bond, yet

\mathrm{NF_3}

(0.23 D) is far less polar than

\mathrm{NH_3}

(1.47 D). The lone-pair moment adds to the bond moments in

\mathrm{NH_3}

and opposes them in

\mathrm{NF_3}

The two arrow conventions point opposite ways

The chemist's arrow runs from positive to negative, following the electron density. The physicist's dipole vector runs from negative to positive. A statement that puts the tail on the negative centre describes the physics convention.

Drill 6 more on dipole moment: size and direction

Concept 2 of 3: Polar or non-polar: when symmetry cancels

Polar bonds do not always make a polar molecule. If identical bonds point symmetrically around the centre with no lone pair to spoil the balance, their moments cancel to zero. Change one outer atom, or add a lone pair that bends the shape, and a net dipole appears.

Definition

Zero dipole: symmetric shapes with identical outer atoms and no net lone-pair moment: linear AX₂, trigonal planar AX₃, tetrahedral AX₄, square planar AX₄E₂, trigonal bipyramidal AX₅, octahedral AX₆, and linear AX₂E₃.
Non-zero dipole: bent, pyramidal, see-saw, T-shaped and square pyramidal shapes; any heteronuclear diatomic; a symmetric shape with mixed outer atoms ( $\mathrm{CHCl_3}$ , $\mathrm{CH_2Cl_2}$ ).
Organic: $p$ -dichlorobenzene and trans-1,2-dichloroethene have zero dipole; the ortho and meta isomers and the cis isomer are polar.
A homonuclear diatomic ( $\mathrm{H_2}$ , $\mathrm{N_2}$ ) is non-polar.

Shape	Net dipole	Examples
Linear AX₂ or AX₂E₃	Zero	$\mathrm{CO_2}$ , $\mathrm{BeF_2}$ , $\mathrm{BeCl_2}$ , $\mathrm{XeF_2}$
Trigonal planar AX₃	Zero	$\mathrm{BF_3}$ , $\mathrm{BCl_3}$ , $\mathrm{SO_3}$
Tetrahedral AX₄	Zero	$\mathrm{CH_4}$ , $\mathrm{CCl_4}$ , $\mathrm{SiF_4}$
Square planar, TBP, octahedral	Zero	$\mathrm{XeF_4}$ , $\mathrm{PCl_5}$ , $\mathrm{SF_6}$
Bent	Non-zero	$\mathrm{H_2O}$ , $\mathrm{H_2S}$ , $\mathrm{SO_2}$
Pyramidal	Non-zero	$\mathrm{NH_3}$ , $\mathrm{NF_3}$ , $\mathrm{PCl_3}$
See-saw, T-shaped, square pyramidal	Non-zero	$\mathrm{SF_4}$ , $\mathrm{ClF_3}$ , $\mathrm{BrF_5}$
Tetrahedral with mixed atoms	Non-zero	$\mathrm{CHCl_3}$ , $\mathrm{CH_2Cl_2}$
Heteronuclear diatomic	Non-zero	HF, HCl, HBr H₂ has zero dipole; HF, with the biggest electronegativity gap, has the largest of the hydrogen halides.

A lone pair on the centre breaks the symmetry unless the lone pairs themselves are placed symmetrically, as in XeF₂ and XeF₄.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q150Moderate

Example 2 · Chemical Bonding and Molecular Structure · Dipole Moment, Hydrogen Bonding and Intermolecular Forces

Number of compounds from the following with zero dipole moment is ____ .

HF,H_{2},H_{2}\text{ }S,CO_{2},NH_{3},BF_{3},CH_{4},CHCl_{3},SiF_{4}

H_{2}O,BeF_{2}

Polar bonds can give a non-polar molecule

C–Cl and B–F bonds are strongly polar, but

\mathrm{CCl_4}

and

\mathrm{BF_3}

have zero dipole moment. Ask whether the shape is symmetric, not whether the bonds are polar.

Lone pairs do not always make a molecule polar

\mathrm{XeF_4}

has two lone pairs and

\mathrm{XeF_2}

has three, yet both are non-polar: the lone pairs sit opposite each other or round the equator, and their moments cancel.

Drill 8 more on polar or non-polar: when symmetry cancels

Concept 3 of 3: Hydrogen bonding and intermolecular forces

A hydrogen bonded to F, O or N is left almost bare, so it is pulled strongly towards a lone pair on a nearby F, O or N. That is a hydrogen bond. It can link two molecules (intermolecular) or two groups inside one molecule (intramolecular), and which one happens changes boiling points and volatility.

Definition

A hydrogen bond needs H covalently bonded to a small, highly electronegative atom (F, O, N) and a lone pair on another such atom.
Intermolecular H-bonds join molecules: water, HF (zig-zag chains), $\mathrm{NH_3}$ , $p$ -nitrophenol. They raise the boiling point.
Intramolecular H-bonds close a ring inside one molecule: $o$ -nitrophenol, salicylaldehyde. The molecule has fewer links to its neighbours, so it boils lower and is steam volatile.
The extent of H-bonding depends on the physical state: ice > liquid water > vapour; dissolved impurities disrupt it.
Other forces: London (dispersion) forces act between all molecules, with energy $\propto 1/r^6$ ; dipole-dipole energy $\propto 1/r^3$ for fixed polar molecules and $1/r^6$ for rotating ones.

Case	Kind of attraction	Effect
HF	Intermolecular H-bonds, zig-zag chains	The strongest single H-bond; the H sits nearer one F, so the bonds are not symmetrical
Ice, water, water with solute	Intermolecular H-bonds	Most in ice (each molecule bonded four ways), fewer in liquid water, fewer again with impurities
$o$ -Nitrophenol, salicylaldehyde	Intramolecular H-bond	Lower boiling point; steam volatile
$p$ -Nitrophenol, $p$ -hydroxybenzaldehyde	Intermolecular H-bonds	Higher boiling point; not steam volatile
$\mathrm{CH_4 < HCN < NH_3}$	None, weak C–H···N, N–H···N	Order of intermolecular H-bond strength
Noble gases, $\mathrm{CH_4}$	London forces only	Energy $\propto 1/r^6$ ; grows with molecular size
Ar, $\mathrm{CH_4}$ , $\mathrm{H_2O}$ , $\mathrm{C_6H_6}$	Van der Waals constant a (about 1.4, 2.3, 5.5, 18 L² bar mol⁻²)	Larger a means stronger attraction between molecules

H bonded to F, O or N gives a hydrogen bond; where it forms, inside or between molecules, decides the boiling point.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q124Moderate

Example 3 · Chemical Bonding and Molecular Structure · Dipole Moment, Hydrogen Bonding and Intermolecular Forces

The correct statement/s about Hydrogen bonding is/are: (A) Hydrogen bonding exists when

H

is covalently bonded to the highly electro negative atom. (B) Intermolecular

H

bonding is present in o-nitro phenol (C) Intramolecular

H

bonding is present in

HF

. (D) The magnitude of

H

bonding depends on the physical state of the compound. (E) H-bonding has powerful effect on the structure and properties of compounds. Choose the correct answer from the options given below:

Ortho means intramolecular

The ortho isomer of a nitrophenol or hydroxybenzaldehyde bonds within itself; the para isomer bonds to its neighbours. Swapping them reverses every boiling-point and volatility answer.

HF has no intramolecular hydrogen bond

One HF molecule has a single H–F bond, so it cannot bond to itself. Its hydrogen bonds link separate molecules into chains, and they are not symmetrical.

Drill 10 more on hydrogen bonding and intermolecular forces

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Dipole moment: size and direction
Dipole moment
$\mu = q \times d \qquad 1\ \text{D} = 10^{-18}\ \text{esu cm}$

Reference tables (2)

Polar or non-polar: when symmetry cancels9 rows

Shape	Net dipole	Examples
Linear AX₂ or AX₂E₃	Zero	$\mathrm{CO_2}$ , $\mathrm{BeF_2}$ , $\mathrm{BeCl_2}$ , $\mathrm{XeF_2}$
Trigonal planar AX₃	Zero	$\mathrm{BF_3}$ , $\mathrm{BCl_3}$ , $\mathrm{SO_3}$
Tetrahedral AX₄	Zero	$\mathrm{CH_4}$ , $\mathrm{CCl_4}$ , $\mathrm{SiF_4}$
Square planar, TBP, octahedral	Zero	$\mathrm{XeF_4}$ , $\mathrm{PCl_5}$ , $\mathrm{SF_6}$
Bent	Non-zero	$\mathrm{H_2O}$ , $\mathrm{H_2S}$ , $\mathrm{SO_2}$
Pyramidal	Non-zero	$\mathrm{NH_3}$ , $\mathrm{NF_3}$ , $\mathrm{PCl_3}$
See-saw, T-shaped, square pyramidal	Non-zero	$\mathrm{SF_4}$ , $\mathrm{ClF_3}$ , $\mathrm{BrF_5}$
Tetrahedral with mixed atoms	Non-zero	$\mathrm{CHCl_3}$ , $\mathrm{CH_2Cl_2}$
Heteronuclear diatomic	Non-zero	HF, HCl, HBr H₂ has zero dipole; HF, with the biggest electronegativity gap, has the largest of the hydrogen halides.

A lone pair on the centre breaks the symmetry unless the lone pairs themselves are placed symmetrically, as in XeF₂ and XeF₄.

Hydrogen bonding and intermolecular forces7 rows

Case	Kind of attraction	Effect
HF	Intermolecular H-bonds, zig-zag chains	The strongest single H-bond; the H sits nearer one F, so the bonds are not symmetrical
Ice, water, water with solute	Intermolecular H-bonds	Most in ice (each molecule bonded four ways), fewer in liquid water, fewer again with impurities
$o$ -Nitrophenol, salicylaldehyde	Intramolecular H-bond	Lower boiling point; steam volatile
$p$ -Nitrophenol, $p$ -hydroxybenzaldehyde	Intermolecular H-bonds	Higher boiling point; not steam volatile
$\mathrm{CH_4 < HCN < NH_3}$	None, weak C–H···N, N–H···N	Order of intermolecular H-bond strength
Noble gases, $\mathrm{CH_4}$	London forces only	Energy $\propto 1/r^6$ ; grows with molecular size
Ar, $\mathrm{CH_4}$ , $\mathrm{H_2O}$ , $\mathrm{C_6H_6}$	Van der Waals constant a (about 1.4, 2.3, 5.5, 18 L² bar mol⁻²)	Larger a means stronger attraction between molecules

H bonded to F, O or N gives a hydrogen bond; where it forms, inside or between molecules, decides the boiling point.

Watch out for (6)

More electronegative F does not mean a bigger dipole
→ Dipole moment: size and direction
The two arrow conventions point opposite ways
→ Dipole moment: size and direction
Polar bonds can give a non-polar molecule
→ Polar or non-polar: when symmetry cancels
Lone pairs do not always make a molecule polar
→ Polar or non-polar: when symmetry cancels
Ortho means intramolecular
→ Hydrogen bonding and intermolecular forces
HF has no intramolecular hydrogen bond
→ Hydrogen bonding and intermolecular forces

Test yourself on Chemical Bonding and Molecular Structure

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