Why Polar Molecules Do Not Always Have the Highest Boiling Point

Suppose you compare pure hydrogen chloride, HCl, with pure hydrogen bromide, HBr. HCl is more polar: its dipole moment is approximately 1.09 D, compared with 0.83 D for HBr. You might therefore predict that HCl has the higher boiling point.

That prediction is wrong.

At normal atmospheric pressure, HCl boils at approximately \(-85.1^\circ\mathrm{C}\), while HBr boils at approximately \(-66.8^\circ\mathrm{C}\). HBr boils about \(18.3^\circ\mathrm{C}\) higher, despite having the smaller permanent dipole.

The reason is that boiling point depends on the combined intermolecular attractions between whole molecules, not polarity alone. HBr has a much larger, more easily distorted electron cloud, producing stronger London dispersion forces. These attractions outweigh the advantage HCl has from its stronger dipole-dipole interactions.

Here, HCl and HBr refer to the pure molecular compounds, not their aqueous acid solutions.

01What actually determines a boiling point?

Before comparing molecules, think about what happens when a liquid boils.

A liquid contains molecules that are relatively close together. Attractions between neighbouring molecules help keep them in the liquid.

When the liquid is heated, its molecules gain kinetic energy. Eventually, enough molecules can escape the liquid and enter the gas phase for boiling to occur.

For a pure substance, boiling occurs when:

\[ \text{Vapour pressure}=\text{External pressure} \]

The vapour pressure is the pressure exerted by the substance’s vapour when it is in equilibrium with its liquid.

At the same temperature, stronger overall intermolecular attractions generally mean fewer molecules escape into the vapour. This produces a lower vapour pressure.

The liquid must therefore reach a higher temperature before its vapour pressure equals the external pressure.

This is why stronger overall intermolecular attractions generally produce higher boiling points when comparing substances at the same external pressure.

The normal boiling point is measured at an external pressure of \(101.325\ \mathrm{kPa}\).

Boiling does not break covalent bonds

There is another distinction worth getting right.

During boiling, molecules separate from one another. The covalent bonds holding the atoms inside each molecule remain intact.

For example:

\[ \ce{HCl(l) -> HCl(g)} \]

The HCl molecules move further apart, but each molecule still contains one hydrogen atom covalently bonded to one chlorine atom.

There are two different types of attraction involved:

If you describe boiling as breaking covalent bonds, you’re describing the wrong process. For a more detailed explanation, see Why Boiling Water Does Not Break Its Covalent Bonds.

02Why HBr boils higher than HCl

Let’s compare their complete molecular structures rather than looking at polarity in isolation.

Both HCl and HBr are diatomic molecules. Each contains a hydrogen atom covalently bonded to a halogen atom.

Both molecules are polar because their atoms have different electronegativities.

However, chlorine and bromine differ considerably in size.

The dipole moments are experimental values from the NIST Computational Chemistry Comparison and Benchmark Database. The boiling-point data are approximate reference values for pure substances at \(101.325\ \mathrm{kPa}\), available through the NIST Chemistry WebBook for HCl and HBr.

1. HCl has stronger dipole-dipole attractions

A dipole moment measures the permanent separation of positive and negative charge within a molecule.

A larger dipole moment generally indicates stronger permanent polarity.

HCl has a dipole moment of approximately 1.093 D, while HBr has a dipole moment of approximately 0.827 D.

Therefore, HCl has the stronger permanent dipole.

Both substances experience dipole-dipole attractions between neighbouring molecules, but HCl has the advantage here.

If dipole-dipole attractions were the only important factor, you would expect HCl to have the higher boiling point.

But they aren’t the only factor.

2. HBr has stronger London dispersion forces

London dispersion forces occur because electrons are constantly moving.

At any instant, an electron cloud can become unevenly distributed. This creates a temporary dipole.

That temporary dipole can distort the electron cloud of a neighbouring molecule, producing an induced dipole. The resulting attraction is a London dispersion force.

These forces exist between all molecules, including non-polar molecules.

The strength of London dispersion forces depends partly on how easily an electron cloud can be distorted.

This property is called polarisability.

Think of comparing two balloons: a larger, softer balloon is easier to distort than a smaller, firmer one. A similar idea helps explain why a larger electron cloud can be more easily distorted.

The analogy has a limit: electron clouds are probability distributions, not physical balloons. It simply helps picture the idea of polarisability.

Now compare the two molecules:

• HCl contains 18 electrons.
• HBr contains 36 electrons.

Bromine has a larger, more diffuse electron cloud than chlorine. HBr is therefore more polarizable and experiences stronger London dispersion forces.

Its stronger dispersion attractions compensate for, and outweigh, its weaker dipole-dipole attractions.

Consequently, more energy is needed to separate HBr molecules sufficiently for boiling.

HBr has the higher boiling point.

3. Why isn’t hydrogen bonding the explanation?

You might notice that both molecules contain hydrogen and wonder whether hydrogen bonding is involved.

It isn’t, under the standard HSC Chemistry model.

For conventional hydrogen bonding, hydrogen must be covalently bonded to nitrogen, oxygen, or fluorine. The other molecule must also provide a suitable lone pair.

In HCl and HBr, hydrogen is bonded to chlorine or bromine instead.

Their important intermolecular attractions are therefore:

• London dispersion forces.
• Dipole-dipole attractions.

The difference in boiling point comes from the balance between these two types of attraction, particularly the stronger dispersion forces in HBr.

03How was the boiling-point difference established experimentally?

So far, we’ve used molecular structure to explain the difference. But how do we know which substance actually boils at the higher temperature?

The experimental question is straightforward:

At the same external pressure, what temperature does each pure substance reach when its liquid and vapour are in equilibrium during boiling?

There is a problem with the simplest experimental design.

HCl and HBr are both gases at room temperature. You cannot simply pour them into open beakers and heat them until they boil.

The first experimental design

Imagine placing each substance in an open test tube, positioning a thermometer near the opening, and recording the temperature.

This approach fails for several reasons:

• Both substances are already gases at room temperature.
• An open container cannot reliably retain a liquid sample.
• Escaping gas does not provide a stable liquid-vapour equilibrium measurement.
• A thermometer near the opening does not necessarily measure the temperature of a liquid sample.

The apparatus cannot produce a reliable boiling-point comparison.

A better measurement arrangement

A more suitable measurement model uses a sealed, pressure-rated equilibrium cell.

The apparatus contains:

1. A sealed, corrosion-resistant cell containing the pure substance.
2. A temperature-controlled bath surrounding the cell.
3. A calibrated temperature sensor measuring the sample temperature.
4. A pressure transducer connected to the vapour space inside the cell.

The sample is first cooled sufficiently to establish liquid-vapour coexistence.

The bath temperature is then increased gradually. As the temperature rises, the substance’s equilibrium vapour pressure increases.

The pressure transducer records the pressure inside the cell while the temperature sensor records the corresponding temperature.

When the vapour pressure reaches \(101.325\ \mathrm{kPa}\), the corresponding temperature is the normal boiling point.

This is a conceptual measurement arrangement, not a suggested school practical. Hydrogen halides are corrosive substances and require specialist containment and equipment.

What should the measurements show?

Consider two competing predictions.

Prediction 1: Polarity alone determines boiling point.

HCl has the larger dipole moment, so this model predicts that HCl should have the higher boiling point.

Prediction 2: The combined intermolecular attractions determine boiling point.

HBr has stronger London dispersion forces because its larger electron cloud is more polarizable. This model predicts that HBr could have the higher boiling point despite its smaller dipole moment.

The published reference data support the second prediction.

HBr boils approximately \(18.3^\circ\mathrm{C}\) higher than HCl.

The observation challenges the idea that the most polar molecule must always have the highest boiling point.

It does not show that dipole-dipole attractions are unimportant. It shows that they must be considered alongside other intermolecular forces.

The NIST values are compiled reference data, not measurements from the conceptual apparatus described above. The apparatus explains the measurement principle and the controls needed for a fair comparison.

04A contrasting case: when polarity does increase the boiling point

It would be equally wrong to conclude that polarity never matters.

Consider two structural isomers: cis-1,2-dichloroethene and trans-1,2-dichloroethene.

Both have the molecular formula \(\ce{C2H2Cl2}\), and both have a molar mass of approximately \(96.95\ \mathrm{g\,mol^{-1}}\).

Their molecular arrangements are different.

In the cis isomer, the chlorine atoms are on the same side of the double bond. The C-Cl bond dipoles do not cancel, so the molecule has a net dipole.

In the trans isomer, the chlorine atoms are on opposite sides. The molecular arrangement is symmetrical, so the bond dipoles cancel.

The comparison is:

The boiling-point ranges and molecular masses are reported in the ATSDR Toxicological Profile for 1,2-Dichloroethene.

Both molecules experience London dispersion forces. Their similar molecular formulas and electron counts mean their dispersion attractions are broadly comparable.

However, the cis molecule also experiences dipole-dipole attractions.

These additional attractions contribute to its higher boiling point.

There is one important limitation: the two isomers also have different geometries, which can affect how neighbouring molecules interact. This is therefore useful evidence for the contribution of polarity, but it is not a perfect experiment in which only polarity changes.

The lesson is not that polarity is irrelevant. It is that its importance depends on the other properties of the molecules being compared.

For more on the effect of molecular geometry and shape, see How Molecular Shape Explains Boiling-Point Differences.

05A reliable method for HSC boiling-point comparisons

When an exam asks you to compare the boiling points of two molecular substances, avoid immediately deciding which one is more polar.

Instead, work through the complete molecular structures.

A useful way to structure your final explanation is:

1. Identify the intermolecular forces present in both substances.
2. Explain which structural feature changes the strength of those forces.
3. Explain which substance has stronger overall intermolecular attractions.
4. Connect those attractions to vapour pressure and boiling point.
5. State that covalent bonds within the molecules remain intact.

This approach is much more reliable than writing that one molecule boils higher simply because it is more polar or has a greater molar mass.

06Worked examples

07Questions and solutions

Question 1

Two hypothetical pure molecular liquids, A and B, have similar molar masses, molecular shapes, and polarisabilities.

Neither substance can form hydrogen bonds.

Molecule A is polar, while molecule B is non-polar.

Predict which substance would generally have the higher boiling point at the same external pressure. Explain your reasoning.

Solution 1

Molecule A would generally have the higher boiling point.

Both substances experience London dispersion forces, and their similar sizes and polarisabilities suggest that these attractions should be broadly comparable.

However, molecule A also experiences dipole-dipole attractions because it has a permanent molecular dipole.

Molecule B has no permanent molecular dipole, so it does not experience permanent dipole-dipole attractions.

The additional intermolecular attractions in A generally make it more difficult for its molecules to separate into the gas phase.

Therefore, A would generally have the higher boiling point.

The important assumption is that their sizes, shapes, and polarisabilities are sufficiently similar. If those properties differed substantially, the result might change.

Question 2

Hydrogen bromide and hydrogen iodide have the following properties:

Both molecules are polar and neither forms conventional hydrogen bonds with itself under the standard HSC model.

(a) Calculate the difference between their normal boiling points.

(b) Explain why HI has the higher boiling point despite having a smaller dipole moment.

(c) Does the observation mean that dipole-dipole attractions are absent in HI? Explain.

Solution 2

(a) HI has a normal boiling point approximately \(31.4^\circ\mathrm{C}\) higher than HBr.

Calculate the difference:

\[ \begin{aligned} \Delta T_b&=T_{b,\mathrm{HI}}-T_{b,\mathrm{HBr}}\\ &=-35.4-(-66.8)\\ &=31.4^\circ\mathrm{C} \end{aligned} \]

(b) HI has stronger London dispersion forces because its electron cloud is larger and more polarizable.

HI contains 54 electrons per molecule, compared with 36 in HBr.

Its larger electron cloud is more easily distorted, producing stronger temporary dipoles and induced dipoles.

Although HBr has the larger permanent dipole moment and therefore stronger dipole-dipole attractions, HI’s stronger dispersion attractions outweigh this difference.

Consequently, more energy is needed to separate HI molecules during boiling.

(c) No. HI still experiences dipole-dipole attractions.

HI is a polar molecule, so neighbouring molecules experience permanent dipole-dipole attractions.

However, the strength of one type of intermolecular force does not determine the overall boiling point by itself.

HI’s stronger dispersion attractions dominate the comparison, giving it the higher boiling point.

Question 3

Pure hydrogen fluoride, HF, has a normal boiling point of approximately \(19.5^\circ\mathrm{C}\), while pure hydrogen chloride, HCl, boils at approximately \(-85.1^\circ\mathrm{C}\).

Their molar masses are:

• HF: \(20.01\ \mathrm{g\,mol^{-1}}\)
• HCl: \(36.46\ \mathrm{g\,mol^{-1}}\)

Explain why HF has the much higher boiling point, even though HCl has the larger molar mass.

Your answer should identify the important intermolecular forces in both substances.

Solution 3

HF has the higher boiling point because its molecules form hydrogen bonds with one another.

Both substances are polar and experience London dispersion and dipole-dipole attractions.

However, HF has an additional type of intermolecular attraction.

Fluorine is highly electronegative, so the H-F bond is strongly polar. The partially positive hydrogen atom in one HF molecule can attract a lone pair on the fluorine atom of another HF molecule.

This produces intermolecular hydrogen bonding.

HCl does not form conventional hydrogen bonds under the standard HSC model because hydrogen is bonded to chlorine rather than fluorine, oxygen, or nitrogen.

Although HCl has a larger electron cloud and stronger dispersion attractions, HF’s hydrogen bonding produces much stronger overall intermolecular attractions.

HF therefore requires a higher temperature to reach its boiling point.

The important lesson is that a smaller molecule can have a higher boiling point when it can form sufficiently strong intermolecular attractions.

Question 4

A laboratory investigation produces the following illustrative measurements:

A student calculates:

\[ -70.5-(-85.0)=14.5^\circ\mathrm{C} \]

The student concludes that HBr’s normal boiling point is exactly \(14.5^\circ\mathrm{C}\) higher than HCl’s.

Is this conclusion correct?

Can we still determine which substance has the higher normal boiling point? Explain.

Solution 4

The calculated difference of \(14.5^\circ\mathrm{C}\) is not the difference between their normal boiling points, but we can still establish that HBr has the higher normal boiling point.

The problem is that the two substances were measured at different external pressures.

Boiling point depends on external pressure because boiling occurs when vapour pressure equals external pressure.

Increasing the external pressure requires a higher temperature before the vapour pressure reaches the required value.

The HBr measurement was taken at only \(80.0\ \mathrm{kPa}\). If its external pressure were increased to \(101.3\ \mathrm{kPa}\), its boiling point would also increase.

Its normal boiling point must therefore be higher than its measured value of \(-70.5^\circ\mathrm{C}\).

This is already higher than HCl’s normal boiling point of \(-85.0^\circ\mathrm{C}\).

We can therefore establish that:

\[ T_{b,\mathrm{HBr}}>T_{b,\mathrm{HCl}} \]

However, the exact difference cannot be obtained from the two measurements provided.

A fair experimental comparison would measure both substances at the same external pressure, using calibrated temperature and pressure sensors and pure samples.

This question highlights an important experimental principle: a difference in measured results is only meaningful when the conditions affecting those results have been properly considered.

Question 5

Two investigations compare the boiling points of different molecular substances.

Investigation A: Hydrogen halides

Investigation B: 1,2-dichloroethene isomers

A student makes two claims:

• Investigation A proves that dipole-dipole attractions do not significantly affect boiling points.
• Investigation B proves that the molecule with the larger dipole moment must always have the higher boiling point.

Evaluate both claims.

Which investigation provides stronger evidence that molecular polarity contributes to boiling-point differences?

Identify an important limitation of that investigation, and propose how a chemist could obtain stronger evidence for the contribution of dipole-dipole attractions.

Solution 5

Neither claim is justified. Investigation B provides stronger evidence for the contribution of dipole-dipole attractions, but it does not prove that polarity alone determines boiling point.

The key is to examine what changes in each investigation.

First, consider Investigation A.

HCl has the larger dipole moment, but HBr has the higher boiling point.

A student might interpret this as evidence that dipole-dipole attractions are unimportant.

However, the comparison changes two important properties simultaneously.

HBr has a substantially larger electron cloud and greater polarisability. Its London dispersion forces are therefore stronger.

The increase in dispersion attractions outweighs the advantage HCl has from its stronger permanent dipole.

Investigation A demonstrates that dipole moment alone cannot predict boiling point. It does not demonstrate that dipole-dipole attractions have no effect.

Now consider Investigation B.

The cis and trans isomers have:

• The same molecular formula.
• The same molar mass.
• The same number of electrons.
• Similar overall molecular sizes.
• Different molecular geometries and overall polarities.

Both experience London dispersion forces.

However, the cis isomer has a net molecular dipole, while the trans isomer’s bond dipoles approximately cancel.

The additional dipole-dipole attractions in the cis isomer are consistent with its higher boiling point.

This makes Investigation B a better comparison for examining the contribution of molecular polarity because molecular size, molar mass, and electron count are held constant.

But there is an important limitation.

The molecular geometry also changes between the two isomers.

This means their shapes and the way neighbouring molecules interact can differ. Their dispersion attractions may therefore not be identical.

The experiment has controlled several important properties, but it has not changed polarity while keeping every other structural feature perfectly constant.

A stronger investigation would compare several carefully selected pairs of molecular substances.

A chemist could:

1. Select molecules with similar molar masses, electron-cloud sizes, and molecular shapes.
2. Exclude molecules that can form hydrogen bonds, so this additional force does not complicate the comparison.
3. Compare molecules with different permanent dipole moments.
4. Measure their vapour pressures or normal boiling points at the same external pressure.
5. Look for consistent relationships across several comparisons rather than relying on one pair.

This would provide stronger evidence for the contribution of dipole-dipole attractions while reducing the influence of other structural differences.

The final conclusion is that boiling point cannot be predicted reliably from polarity alone. Molecular size, polarisability, geometry, and hydrogen bonding all contribute to the intermolecular attractions between whole molecules.

That is why the next step in a boiling-point comparison should always be to examine the complete molecular structure, including its three-dimensional shape, rather than stopping after identifying whether a molecule is polar.