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Chemical Bonding and Molecular Structure · TOPIC LESSON

Intermolecular Forces

Relate molecular shape, dispersion, polarity and hydrogen bonding to phase and liquid properties.

JEE Main · JEE Advanced · NEET UG46 min study guide9 indexed sections

BY THE END, YOU CAN

  • Distinguish intermolecular separation from covalent bond breaking.
  • Explain dispersion, permanent and induced dipoles, and ion–dipole orientation.
  • Identify hydrogen-bond donors, acceptors and internal versus external contacts.
  • Apply vaporization and surface-energy formulas with explicit units.
  • Use property trends without ignoring packing, pressure or competing forces.

BEFORE YOU START

  • Lewis structures, geometry and molecular polarity.
  • Basic thermodynamics and phase changes.
  • Moles, energy units and simple powers.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. First decide what is being separated

Intermolecular interactions act between molecules, while intramolecular bonds hold atoms together within a molecule. Boiling liquid water mostly separates H₂O molecules from one another; it does not ordinarily split each molecule into hydrogen and oxygen atoms. The energy scale and products of vaporization therefore differ from those of covalent bond dissociation.

The phrase intermolecular forces is sometimes used broadly for interactions among particles including ions and noble-gas atoms. Be precise about the system: a molecular liquid, an ionic crystal and a metallic solid do not have the same microscopic bonding picture. Diamond’s high melting or decomposition temperature cannot be explained as merely strong attractions between discrete carbon molecules.

Attractions favour close association, while thermal motion and entropy favour accessible configurations, including dispersed states. Phase behaviour depends on temperature and pressure as well as the interaction potential. Stronger attractions do not guarantee that a substance is solid under every condition.

A molecule can experience several types of interaction at once. Polar molecules still have dispersion interactions, and hydrogen-bonding molecules still have both dispersion and other electrostatic contributions. Identifying a dominant contribution is not permission to erase the others.

ORIGINAL PRACTICE · 1

When molecular iodine sublimes, must every I–I covalent bond break?

Reveal answer & reasoning

No.

I₂ molecules separate from the molecular solid; sublimation differs from dissociation to iodine atoms.

02 / UNDERSTAND & APPLY

2. Dispersion exists even without a permanent dipole

London dispersion arises from correlated fluctuations in electron distributions. A temporary fluctuation in one particle is associated with an induced response in another, producing an attractive contribution. An average permanent dipole of zero does not imply that electrons are motionless or that the particle has no attractive interactions.

Polarizability describes how readily an electron cloud is distorted. Larger, more diffuse clouds are often more polarizable; comparisons within a related series frequently show stronger dispersion for heavier members. Electron count and size are useful clues, but molar mass alone is not a universal interaction-strength law.

Molecular shape matters. Extended isomers can make more extensive close contact than compact branched isomers, often increasing boiling point within a comparable hydrocarbon family. Crystal packing can produce a different melting-point order, so a boiling trend should not be copied mechanically into a melting trend.

At sufficiently long range, a simple nonretarded pair dispersion potential varies approximately as −C₆/r⁶. The negative sign means attraction relative to separated particles. At very short distances electron-cloud overlap gives strong repulsion, so the attractive term alone cannot describe the entire interaction or a stable separation.

ORIGINAL WORKED EXAMPLE

A long-range model ratio

In a stated pair model U_disp=−C₆/r⁶, separation increases from r to 2r. What happens to the magnitude of the attractive energy?

  1. At r, magnitude is C₆/r⁶.
  2. At 2r, magnitude is C₆/(2r)⁶=(1/64)C₆/r⁶.
  3. The energy remains negative but is closer to zero; this is a long-range pair-model result.

Answer The magnitude falls to 1/64 of its original value.

ORIGINAL PRACTICE · 1

Why can nonpolar argon atoms condense at sufficiently low temperature?

Reveal answer & reasoning

They experience dispersion attractions.

A permanent dipole is not required for correlated electron fluctuations and attractive interactions.

03 / UNDERSTAND & APPLY

3. Charge distribution and molecular geometry work together

A permanent molecular dipole is the vector sum of charge-separation contributions over the molecule. CO₂ has polar C=O bonds but a linear symmetric arrangement, giving zero net molecular dipole. Bent H₂O has a nonzero molecular dipole. Counting polar bonds without geometry is insufficient.

Permanent dipoles interact in an orientation-dependent way. Favourable orientations place opposite partial charges nearer each other; other orientations can be less favourable or repulsive. Thermal rotation means a liquid contains a distribution of orientations, not a static perfectly aligned set of molecular arrows.

A polar molecule can also induce a dipole in a nearby polarizable nonpolar molecule. Ion–dipole interactions occur between a full ionic charge and a molecular dipole, as in hydration. Around a cation, water tends to orient its oxygen-rich negative end inward; around an anion, its hydrogen-rich positive side points toward the ion.

A common hierarchy such as ion–dipole stronger than ordinary dipole–dipole can be useful for comparable cases, but it is not an absolute ranking across all molecule sizes and separations. A large highly polarizable nonpolar molecule can have stronger overall cohesion than a much smaller polar molecule.

ORIGINAL WORKED EXAMPLE

Use geometry before assigning dipole attraction

Compare CO₂ and H₂O for permanent molecular dipole, explaining why polar bonds alone do not settle the question.

  1. CO₂ is linear with equal opposing bond dipoles, so their vector sum is zero.
  2. H₂O is bent, so its bond-dipole contributions do not cancel.
  3. Both still have dispersion interactions, while water also has permanent-dipole and hydrogen-bonding contributions.

Answer CO₂ has no permanent net dipole in its symmetric geometry; H₂O does.

ORIGINAL PRACTICE · 1

Which end of a water molecule tends to point toward Na⁺ in its hydration environment?

Reveal answer & reasoning

The oxygen end.

Oxygen carries the more negative region of water’s charge distribution and is attracted to the cation.

04 / UNDERSTAND & APPLY

4. Identify a donor, an acceptor and the geometry

A hydrogen bond involves an X–H donor group and an electron-rich acceptor. In common school examples X is N, O or F, and the acceptor is often a suitable lone pair on N, O or F. This is a useful recognition rule, not a complete modern definition excluding every weaker C–H or other example.

The covalent X–H bond is not itself the hydrogen bond. Write X–H···Y, with the dotted contact between H and acceptor Y. Hydrogen bonding is directional, often favouring near-linear donor–H–acceptor alignment, and includes electrostatic, polarization and other electronic contributions rather than a single rigid ionic model.

An alcohol can donate through O–H and accept through oxygen. An ether can accept from water or alcohol but lacks an O–H/N–H/F–H donor for ordinary strong self-hydrogen bonding. A carbonyl oxygen is an acceptor; an amide nitrogen’s lone pair is delocalized and is not normally a hydrogen-bond acceptor, although an amide N–H can donate.

Water has two donor hydrogens and approximately two acceptor directions, supporting an extended network. A water molecule can have four hydrogen-bond contacts in an ideal tetrahedral network, but liquid contacts continually break and reform. Counting contacts per molecule is different from counting distinct bonds in the whole network, because each bond is shared.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Methanol donates a hydrogen bond to an ether oxygenMethanol CH3-O-H is drawn with a dashed contact from its hydroxyl hydrogen to the oxygen of dimethyl ether. The ether oxygen has two covalent bonds to methyl groups. The solid donor O-H bond is distinguished from the dashed intermolecular H-to-O contact.A donor and an acceptor can be different moleculesCH₃OHOCH₃CH₃MethanolDimethyl etherO–H donorO acceptorDashed H···O: hydrogen bond between moleculesSolid O–H: covalent bond within methanolEther accepts here but has no O–H donor of its own.
Connectivity and contact schematic; complete lone-pair distributions and bond angles are omitted. The ether acts as acceptor, not donor.

ORIGINAL WORKED EXAMPLE

Donor/acceptor classification

Classify methanol and dimethyl ether as ordinary hydrogen-bond donors and acceptors, and decide whether ether can hydrogen-bond with water.

  1. Methanol has O–H and oxygen lone pairs, so it can donate and accept.
  2. Dimethyl ether has oxygen lone pairs but no O–H donor, so it accepts but does not donate in the usual strong hydrogen-bond classification.
  3. Water supplies a donor, so water–ether hydrogen bonding is possible.

Answer Methanol: donor and acceptor. Ether: acceptor only in this model; it can hydrogen-bond with water.

ORIGINAL PRACTICE · 1

Does merely containing H and O guarantee that a pure compound can strongly self-hydrogen-bond?

Reveal answer & reasoning

No.

A suitable donor bond and acceptor are needed; an ether contains H and O but has no O–H donor.

ORIGINAL PRACTICE · 2

An ideal periodic water network has N molecules, each with four hydrogen-bond contacts. How many distinct hydrogen bonds are counted?

Reveal answer & reasoning

2N.

Four contacts per molecule gives 4N ends of bonds; divide by two because each bond connects two molecules.

05 / UNDERSTAND & APPLY

5. Internal and intermolecular hydrogen bonds compete

Intermolecular hydrogen bonding links different molecules, whereas intramolecular hydrogen bonding connects a donor and acceptor within the same molecule. The latter requires a suitable spatial arrangement; simply having both groups somewhere in a formula is not enough. Five- or six-membered pseudo-ring arrangements are common favourable motifs.

An internal contact can stabilize a conformation and reduce the availability of a donor or acceptor for intermolecular association. It may thereby affect volatility and solubility, but it does not eliminate every interaction with neighbouring molecules. Solvent and temperature can shift the balance between internal and external contacts.

Ortho-nitrophenol is a familiar example in which neighbouring phenolic and nitro groups permit an internal O–H···O contact. The para isomer cannot make the same short internal loop and more readily supports intermolecular association in the usual comparison. These structural tendencies help explain qualitative separation and property differences; exact physical values still depend on phase and conditions.

Hydrogen bonding in proteins and nucleic acids also depends on donor/acceptor arrangement. A local internal contact can help organize one molecule, while complementary contacts between chains support assemblies. Calling every hydrogen bond intermolecular would miss this distinction.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Intramolecular hydrogen bond in an ortho-nitrophenol arrangementA benzene ring bears adjacent hydroxyl and nitro groups. The hydroxyl hydrogen points toward one nitro oxygen with a dashed hydrogen-bond contact, forming a short internal loop within one molecule. The nitro group is drawn as one conventional resonance structure with N-plus, one double-bonded oxygen and one single-bonded O-minus.An internal contact requires neighbouring groupsOrtho-nitrophenol: one possible local arrangementOHON⁺O⁻Dashed O–H···O contact lies within one molecule.The para isomer cannot form this same short internal loop.One nitro resonance form; full geometry and solvent effects omitted.
Illustrative local conformation rather than measured geometry. Intramolecular hydrogen bonding can compete with solvent and intermolecular contacts; it does not eliminate all external interactions.

ORIGINAL PRACTICE · 1

Why is intramolecular hydrogen bonding a conformational question rather than just a count of heteroatoms?

Reveal answer & reasoning

The donor and acceptor must approach with suitable distance and orientation within one molecule.

A formula can contain both groups yet place them too far apart or poorly aligned.

06 / UNDERSTAND & APPLY

6. Boiling and vapour pressure compare phases

At a fixed temperature, stronger overall liquid cohesion often corresponds to lower equilibrium vapour pressure for comparable molecular substances. Boiling occurs when vapour pressure equals the external pressure, so lower external pressure lowers the boiling temperature without changing the molecule’s covalent formula.

Normal boiling point refers to a specified standard external pressure convention, commonly 1 atm in introductory use. Comparing boiling temperatures measured at different pressures without correction can obscure the intermolecular trend. Hydrogen-bond donors such as small alcohols often boil above similarly sized ethers, but size and other interactions must also be considered.

Vaporization enthalpy is positive because energy is required overall to move molecules from liquid to vapour. For an isothermal phase change at the stated pressure, q=nΔH_vap. Heating the liquid before boiling or heating the vapour afterward requires separate heat-capacity terms.

The approximate Clausius–Clapeyron relation ln(P₂/P₁)=−ΔH_vap/R(1/T₂−1/T₁) assumes vapour behaves approximately ideally, liquid volume is negligible and ΔH_vap is nearly constant over the interval. Temperatures must be kelvin and energy units must match R.

ORIGINAL WORKED EXAMPLE

Separate heating from vaporization

A model liquid already at its boiling temperature has ΔH_vap=32.0 kJ mol⁻¹. What heat is needed to vaporize 0.250 mol at the stated pressure, with no heat loss?

  1. The sample is already at the phase-change temperature, so no initial warming term is required.
  2. q=nΔH_vap=0.250×32.0=8.00 kJ.
  3. This separates molecules and does not imply breaking every covalent bond inside them.

Answer 8.00 kJ absorbed.

ORIGINAL PRACTICE · 1

Why should boiling point decrease when external pressure decreases?

Reveal answer & reasoning

A lower temperature is sufficient for vapour pressure to equal the lower external pressure.

The equilibrium boiling condition changes; the covalent molecule need not change.

07 / UNDERSTAND & APPLY

7. Cohesion affects flow and surfaces, but not through one universal number

Viscosity is resistance to flow. Intermolecular interactions, chain entanglement, shape and temperature all affect how molecules move past one another. Stronger association can increase viscosity within a suitable comparison, while warming many ordinary liquids reduces viscosity. It is not possible to rank every liquid by a single hydrogen-bond count.

Surface tension is energy per increase in surface area, equivalently force per length, with units J m⁻²=N m⁻¹. Molecules at an interface have a different environment from molecules in the bulk. Creating more interface costs free energy under the specified conditions, which helps explain why small free droplets tend toward low-area shapes.

Cohesion refers to interactions within a material, while adhesion concerns contact with another material. Wetting and capillary behaviour depend on both and on geometry. High surface tension does not by itself guarantee that a liquid will rise in every capillary surface.

Solubility requires a free-energy balance of solute–solute, solvent–solvent and solute–solvent changes plus entropy. “Like dissolves like” is a useful guide, not a proof. Small alcohols can interact well with water, whereas extending their nonpolar chain can reduce water compatibility even though an OH group remains.

ORIGINAL WORKED EXAMPLE

Surface work with the area specified

A model interface has constant γ=0.050 N m⁻¹. What reversible work is needed to increase its total interfacial area by 0.020 m²?

  1. Use ΔG_surface=γΔA for the stated constant-tension model.
  2. 0.050 N m⁻¹×0.020 m²=0.0010 N m=0.0010 J.
  3. The problem supplies total area; a film with two surfaces would require including both if only one face area were given.

Answer 1.0×10⁻³ J, or 1.0 mJ.

ORIGINAL PRACTICE · 1

Why can a long-chain alcohol be less water-soluble than a short-chain alcohol despite both having an OH group?

Reveal answer & reasoning

Its larger nonpolar region changes the total free-energy balance of mixing.

One hydrogen-bonding group does not override the rest of the molecular structure.

08 / UNDERSTAND & APPLY

8. Water’s network explains anomalies, not every property by itself

The open hydrogen-bonded arrangement of ordinary ice is less dense than liquid water near freezing, allowing ice to float. Melting partly collapses this open network even though it also increases thermal motion. This is a structural explanation, not a claim that heating normally makes every substance denser.

Water’s relatively high boiling point and vaporization enthalpy for its small molecular size reflect substantial association. Its liquid network is dynamic, so it should not be drawn as one frozen tetrahedral lattice that never changes. Pressure and temperature can produce other ice structures beyond the ordinary school comparison.

When ranking molecular properties, first control what is comparable: composition, size, shape, polarity, donor/acceptor ability and measured conditions. Then identify competing effects. A strong single hydrogen bond need not produce the highest boiling point if the molecule forms few contacts or the comparison has a much larger dispersion contribution.

Melting additionally depends strongly on crystal packing and symmetry. Branching may lower a hydrocarbon boiling point yet improve crystal packing and raise its melting point. The observable must be named before an intermolecular explanation is selected.

09 / UNDERSTAND & APPLY

9. A structure-to-property checklist

Draw or infer geometry, determine whether a permanent dipole survives vector cancellation, identify polarizable regions and locate real hydrogen-bond donors and acceptors. For mixtures, inspect cross-interactions rather than only the pure substances.

State the process and its conditions: vaporization, melting, flow, surface creation, hydration or dissolution. Use a formula only after identifying its assumptions and units. A phase-change energy is not a covalent bond dissociation energy, and a qualitative trend is not an exact numerical prediction.

The examples and practice are original. They emphasize competing interactions and limits so that a familiar phrase such as hydrogen bonding becomes an explanation rather than a catch-all answer.

KEEP THE BIG IDEAS

Quick revision

  • Dispersion is present even without a permanent dipole.
  • Hydrogen bonding requires appropriate donor/acceptor geometry.
  • Internal and intermolecular contacts can compete.
  • Pressure and phase matter in boiling comparisons.
  • A single interaction label cannot rank every physical property.
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The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

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