Polarity — Concepts, Formulas & Examples

Polarity describes how electron density is distributed in a molecule. CBSE Class 11 and NEET test bond polarity, molecular dipole and the ‘like dissolves like’ rule.

Core Concepts

Bond polarity

A bond is polar if the two atoms differ in electronegativity. The more electronegative atom bears a partial negative charge ( $\delta^-$ ), the other a partial positive ( $\delta^+$ ). H-Cl is polar; H-H is not.

The greater the electronegativity difference ( $\Delta\text{EN}$ ), the more polar the bond. As a rough guide:

$\Delta\text{EN} = 0$ : non-polar covalent (H $_2$ , Cl $_2$ )
$0 < \Delta\text{EN} < 1.7$ : polar covalent (H-Cl, O-H)
$\Delta\text{EN} > 1.7$ : ionic (NaCl, KBr)

These boundaries are approximate — the transition from polar covalent to ionic is gradual, not a sharp cutoff.

Molecular polarity

A molecule is polar if it has net dipole moment. Depends on both bond polarity and molecular geometry. Symmetric molecules with polar bonds can be non-polar overall (CO $_2$ , CCl $_4$ ).

The rule: Individual bond dipoles are vectors. If the vectors cancel (due to symmetry), the molecule is non-polar. If they do not cancel, the molecule is polar.

Molecule	Geometry	Polar bonds?	Net dipole?
CO $_2$	Linear	Yes	No (vectors cancel)
H $_2$ O	Bent	Yes	Yes (vectors add)
CCl $_4$	Tetrahedral	Yes	No (symmetric)
CHCl $_3$	Tetrahedral	Yes	Yes (not symmetric)
BF $_3$	Trigonal planar	Yes	No (symmetric)
NH $_3$	Trigonal pyramidal	Yes	Yes (lone pair contributes)

Dipole moment

\mu = q \times d

Where $\mu$ = dipole moment (in Debye, D), $q$ = magnitude of charge (in esu), $d$ = distance between charges (in cm). 1 Debye = $3.336 \times 10^{-30}$ C $\cdot$ m.

Product of charge and distance. Measured in Debye units. Water has a dipole moment of 1.85 D; CO $_2$ has zero.

For molecules with multiple polar bonds, the net dipole moment is the vector sum of all individual bond dipoles. We can use the parallelogram law to calculate it.

\mu_{net} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}

Where $\theta$ is the angle between the two bond dipole vectors. For two equal bond dipoles ( $\mu_1 = \mu_2 = \mu$ ):

\mu_{net} = 2\mu\cos\left(\frac{\theta}{2}\right)

Effect on properties

Polar molecules have higher boiling points than non-polar of similar size (due to dipole-dipole interactions). Polar molecules dissolve in polar solvents; non-polar in non-polar. ‘Like dissolves like’.

Intermolecular forces in order of strength:

Ion-dipole (strongest among non-covalent)
Hydrogen bonding (special strong dipole-dipole)
Dipole-dipole
Dipole-induced dipole
London dispersion forces (present in all molecules)

A non-polar molecule like I $_2$ is barely soluble in water but dissolves readily in CCl $_4$ . A polar molecule like sugar dissolves in water but not in hexane. This is the ‘like dissolves like’ principle at work.

Hydrogen bonding

A special case of dipole-dipole. When H is attached to N, O or F, the bond is highly polar, giving strong attraction to lone pairs on other electronegative atoms. Explains water’s anomalously high boiling point.

Two types of hydrogen bonding:

Intermolecular: Between different molecules. Water, alcohols, carboxylic acids. Responsible for high boiling points.
Intramolecular: Within the same molecule. Ortho-nitrophenol has intramolecular H-bonding (between OH and NO $_2$ ), which reduces its boiling point compared to para-nitrophenol (which has intermolecular H-bonding).

This distinction is a favourite NEET question — ortho-nitrophenol is steam-volatile while para-nitrophenol is not, because intramolecular H-bonding makes ortho behave more like a non-polar molecule.

Fajan’s rules and polarisation

When a cation distorts the electron cloud of an anion, the bond gains some covalent character — this is polarisation. Fajan’s rules predict when this happens:

Small cation with high charge: strong polarising power
Large anion with high charge: easily polarised
Cation with non-noble gas configuration (e.g., $\text{Cu}^+$ , $\text{Ag}^+$ ): stronger polarising power

Example: LiCl has more covalent character than NaCl because Li $^+$ is smaller and polarises Cl $^-$ more effectively.

Worked Examples

Both have polar bonds. CO $_2$ is linear — the two C=O dipoles point in opposite directions and cancel. H $_2$ O is bent — the two O-H dipoles add up to give a net dipole.

Water is polar, so it solvates polar and ionic substances by dipole-ion interactions. Oil is non-polar — no favourable interaction with water, so oil and water separate.

The O-H bond dipole is about 1.5 D. The bond angle in water is 104.5°. Using the formula for two equal bond dipoles:

\mu_{net} = 2 \times 1.5 \times \cos\left(\frac{104.5°}{2}\right) = 3.0 \times \cos(52.25°) = 3.0 \times 0.613 = 1.84\;\text{D}

The experimental value is 1.85 D — our calculation matches well.

Both have trigonal pyramidal geometry. In NH $_3$ , the bond dipoles (pointing from H to N) and the lone pair dipole (pointing away from N) are in the same direction — they add up to give a large net dipole (1.47 D).

In NF $_3$ , the bond dipoles point from N to F (fluorine is more electronegative), which is opposite to the lone pair direction. They partially cancel, giving a much smaller net dipole (0.24 D).

Iodine (I $_2$ ) is non-polar. It dissolves well in CCl $_4$ (non-polar) but barely in water (polar). However, if we add KI to water, I $_2$ dissolves because it forms the polyiodide ion I $_3^-$ , which is ionic and water-soluble. This is why we use KI solution (not pure water) to dissolve iodine in the lab.

Solved Problems (Exam Style)

Problem 1 (JEE Main pattern): Arrange the following in order of increasing dipole moment: CH $_4$ , CHCl $_3$ , CCl $_4$ , CH $_2$ Cl $_2$ .

CH $_4$ : all four bonds identical, symmetric tetrahedral. $\mu = 0$ .

CCl $_4$ : all four bonds identical (C-Cl), symmetric tetrahedral. $\mu = 0$ .

CHCl $_3$ : three C-Cl bonds and one C-H bond. The resultant of three C-Cl dipoles points along the C-H bond direction. Net dipole = 1.01 D.

CH $_2$ Cl $_2$ : two C-Cl bonds and two C-H bonds. The C-Cl resultant and C-H resultant point in similar directions. Net dipole = 1.60 D.

Order: CH $_4$ = CCl $_4$ < CHCl $_3$ < CH $_2$ Cl $_2$

Problem 2 (NEET pattern): Which has a higher boiling point — ortho-nitrophenol or para-nitrophenol? Explain.

Ortho-nitrophenol has intramolecular hydrogen bonding between the OH group and the adjacent NO $_2$ group. Molecules do not associate with each other.

Para-nitrophenol has intermolecular hydrogen bonding (OH groups of different molecules interact). This creates stronger intermolecular forces.

Para-nitrophenol has the higher boiling point because intermolecular forces are stronger than intramolecular ones.

Common Mistakes

Saying all polar bonds give polar molecules. Geometry matters.

Confusing dipole moment with charge. Dipole moment is charge times distance.

Writing that non-polar molecules do not interact. They do — via London dispersion forces.

Forgetting the lone pair contribution to dipole moment. In NH $_3$ , the lone pair adds to the net dipole. In NF $_3$ , it opposes the bond dipoles. This is why NF $_3$ has a much lower dipole moment than NH $_3$ despite having more polar bonds.

Assuming that higher electronegativity difference always means higher dipole moment. Dipole moment also depends on bond length. HF has a smaller dipole moment (1.82 D) than HCl might suggest based on $\Delta$ EN alone, because the H-F bond is very short.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2023 and 2024 both had questions on arranging molecules by dipole moment. NEET 2022 tested the ortho vs para nitrophenol boiling point question. CBSE boards frequently ask a three-mark question on hydrogen bonding and its effect on boiling points.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Two rules — polarity depends on both bond and geometry, and like dissolves like. These handle most polarity questions.

Practice Questions

Q1. Why does BF $_3$ have zero dipole moment despite having polar B-F bonds?

BF $_3$ is trigonal planar with 120° angles. The three equal B-F bond dipoles are symmetrically arranged in a plane. Their vector sum is zero. Symmetry cancels the individual dipoles.

Q2. Between HF and HCl, which has a higher boiling point and why?

HF has a much higher boiling point (19.5°C vs -85°C for HCl). Although HCl has a higher molecular weight, HF forms strong hydrogen bonds (H bonded to the highly electronegative F). HCl cannot form hydrogen bonds — it only has dipole-dipole and London forces.

Q3. Calculate the resultant dipole moment of two equal bond dipoles of 1.2 D each at an angle of 120°.

Using $\mu_{net} = 2\mu\cos(\theta/2) = 2 \times 1.2 \times \cos(60°) = 2.4 \times 0.5 = 1.2$ D.

When the angle between two equal dipoles is 120°, the resultant equals each individual dipole.

Q4. Why is ortho-nitrophenol steam-volatile but para-nitrophenol is not?

Ortho-nitrophenol has intramolecular H-bonding (the OH and NO $_2$ groups are adjacent), so it does not form strong intermolecular associations. It behaves like a non-polar molecule and is easily carried by steam. Para-nitrophenol has intermolecular H-bonding, which holds molecules together tightly, making it non-volatile.

Q5. Arrange in order of increasing covalent character: NaCl, MgCl $_2$ , AlCl $_3$ .

By Fajan’s rules, smaller and more highly charged cations polarise the anion more, increasing covalent character. Na $^+$ (large, +1) < Mg $^{2+}$ (smaller, +2) < Al $^{3+}$ (smallest, +3). So covalent character: NaCl < MgCl $_2$ < AlCl $_3$ .

FAQs

Can a molecule with no polar bonds still be polar? No. If there are no polar bonds, there are no bond dipoles to produce a net dipole. A molecule needs at least one polar bond to be polar. However, having polar bonds is necessary but not sufficient — the geometry must also be asymmetric.

Why is CO $_2$ non-polar but SO $_2$ polar? Both have two double bonds to oxygen. CO $_2$ is linear (180°), so the dipoles cancel perfectly. SO $_2$ is bent (~119°) because of a lone pair on sulfur, so the dipoles do not cancel.

What is the unit of dipole moment? The SI unit is coulomb-metre (C $\cdot$ m), but chemists commonly use the Debye (D). 1 D = $3.336 \times 10^{-30}$ C $\cdot$ m. Typical molecular dipole moments range from 0 to about 11 D.

Does dipole moment affect reaction rates? Indirectly, yes. Polar solvents stabilise polar transition states, speeding up reactions that go through polar intermediates (like SN1 reactions). Non-polar solvents favour reactions with non-polar transition states.

Polarity is the concept that connects structure to solubility, boiling point and reactivity. One property, many consequences.