Did you know? Ammonia is about half as heavy as phosphine by molecular mass, yet it boils more than 50°C higher. Strong intermolecular hydrogen bonding in ammonia, rather than molecular size alone, explains the difference.
This original Hougang JC1 H1 Chemistry tutorial uses the 2027 H1 8873 chemical-bonding outcomes to explain NH₃ versus PH₃: N–H polarity, lone pairs, hydrogen bonds, dispersion forces, molecular shape and published normal boiling points.
EDUKATESG · HOUGANG · JC1 H1 CHEMISTRY
14 chapters · Boiling-point evidence · Transfer questions · FAQs · Scientific sources
Complete chapter index: why heavier need not mean higher boiling
- 1. Start with an observation that overturns a mass-only rule
- 2. Calculate the two molecular masses correctly
- 3. Distinguish bonds inside molecules from attractions between molecules
- 4. Why the N–H group supports conventional hydrogen bonding
- 5. Why the PH₃ lone pair does not make it ammonia-like
- 6. Draw the trigonal pyramidal molecular shapes
- 7. Know the normal boiling-point reference rather than guessing from room temperature
- 8. Explain the usual London dispersion trend without denying the exception
- 9. Use arsine as a second control for the mass trend
- 10. Why a boiling point is not a direct molecular-bond-energy measurement
- 11. Distinguish molecular polarity from a guaranteed hydrogen bond
- 12. Transfer the idea to water and hydrogen fluoride
- 13. Repair the four most common wrong explanations
- 14. Final unseen data task: choose the relevant force first
1. Start with an observation that overturns a mass-only rule
Ammonia, NH₃, has a relative molecular mass of about 17, while phosphine, PH₃, has a relative molecular mass of about 34. Yet ammonia boils at approximately −33.3°C under ordinary atmospheric-pressure reference conditions, whereas phosphine boils at about −87.7°C. The smaller molecule has the substantially higher boiling temperature. This is a clue that molecular mass alone cannot explain the forces holding the liquid molecules together.
An H1 student might expect the heavier molecule to boil at a higher temperature because larger electron clouds commonly give stronger London dispersion attractions. That trend is useful when comparing broadly similar non-hydrogen-bonding molecules, but it is not the only intermolecular interaction that can contribute to a boiling point.
Ammonia has a network of significant N–H···N hydrogen bonding between molecules, whereas ordinary phosphine has little of the strong conventional hydrogen bonding characteristic of nitrogen, oxygen and fluorine hydrides. The additional attraction raises the energy required to separate neighbouring ammonia molecules into the gas phase.
The 2027 H1 Chemistry 8873 Chemical Bonding outcome asks students to explain physical properties using hydrogen bonding and other intermolecular forces. The task is a causal comparison, not memorising one isolated pair of negative Celsius temperatures.
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2. Calculate the two molecular masses correctly
Ammonia contains one nitrogen atom and three hydrogen atoms: 14+3(1)=17 in the familiar relative atomic-mass approximation. Phosphine contains one phosphorus atom and three hydrogens: 31+3(1)=34. The phosphine molecule is roughly twice as massive as ammonia in these rounded comparisons.
That difference can influence dispersion interactions, but it is not a calculation of the boiling point. A heavier molecule may have a more polarizable electron cloud and greater instantaneous dipoles; the corresponding London forces depend on electron distribution, shape and how molecules can approach, not on a universal one-to-one mass formula.
A correct student answer should place relative molecular mass in one part of the explanation and the actual additional hydrogen-bonding interaction in another. Giving only the mass numbers does not account for ammonia’s higher boiling temperature.
One helpful check asks students to predict the *usual* dispersion trend first, then explain why this particular pair departs from it. A result becomes memorable when the reason for the exception is clear.
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3. Distinguish bonds inside molecules from attractions between molecules
Within NH₃, nitrogen and hydrogen are connected by covalent N–H bonds. Within PH₃, phosphorus and hydrogen are connected by covalent P–H bonds. The boiling process does not normally require breaking all those covalent bonds and converting the substance into isolated nitrogen, phosphorus and hydrogen atoms.
Boiling largely separates intact molecules from the liquid into the vapour phase. Its temperature depends on the free-energy balance between liquid and gas and is strongly influenced by intermolecular attractions, not just the energies of the covalent N–H and P–H bonds.
Students sometimes say ammonia boils higher because the N–H covalent bond itself is stronger than P–H, with no explanation of attraction between one molecule and another. That confuses intramolecular bonding with intermolecular hydrogen bonding. The latter is the central causal difference in the school comparison.
A tutor can draw two ammonia molecules and mark the dotted intermolecular N–H···N interaction while keeping the solid-line covalent N–H bonds within each molecule distinct.
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4. Why the N–H group supports conventional hydrogen bonding
Hydrogen bonding is a comparatively strong and directional intermolecular attraction that arises under suitable conditions when hydrogen is covalently bonded to a highly electronegative atom such as nitrogen, oxygen or fluorine and interacts with an appropriate lone-pair-bearing acceptor in another molecular environment.
In ammonia, nitrogen is significantly more electronegative than hydrogen, so N–H bonds are polar. Nitrogen also carries a lone pair. A hydrogen atom attached to the N of one molecule can interact with the lone-pair region of N on a neighbouring molecule, producing N–H···N hydrogen bonding.
The dots in N–H···N illustrate an intermolecular interaction, not a normal additional covalent N–N bond. Hydrogen bonding does not mean that liquid ammonia has transformed into one giant ionic lattice or that each molecule permanently bonds to exactly four neighbours in a rigid way.
Hydrogen-bond arrangements change dynamically in the liquid. A school diagram of one donor and one acceptor gives the local interaction responsible for the enhanced cohesion without claiming that the liquid has one perfect permanent molecular arrangement.
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5. Why the PH₃ lone pair does not make it ammonia-like
Phosphine is also a group-15 hydride with a nonbonding electron pair in its Lewis structure, but its P–H bonds have much less polarity than NH₃’s N–H bonds. Phosphorus is not sufficiently electronegative in this context to produce the strong conventional donor behaviour characteristic of N–H, O–H and F–H groups.
Therefore simply counting one lone pair on both molecules is not enough. Hydrogen bonding requires appropriate donor bond polarization, acceptor properties and geometry. Phosphine’s condensed phase is described mainly by dispersion and other relatively weak intermolecular interactions instead of a strong ammonia-like hydrogen-bond network.
It would be overconfident to claim that phosphine can experience no attractive interaction whatsoever with any molecule. All molecules experience dispersion forces, and specialised chemical contexts can involve additional interactions. The useful H1 conclusion is comparative: ordinary pure PH₃ does not exhibit the strong intermolecular hydrogen bonding responsible for NH₃’s unexpectedly high boiling point.
A tutor should ask a student to identify both the donor bond and the lone-pair acceptor, not merely circle the word lone pair in a dot-and-cross diagram.
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6. Draw the trigonal pyramidal molecular shapes
Both NH₃ and PH₃ have central group-15 atoms bonded to three hydrogen atoms, with one nonbonding electron pair in the simple VSEPR model. Four electron regions give a tetrahedral electron-pair arrangement, but the molecular shape is described as trigonal pyramidal because a lone pair occupies one region.
That same broad shape does not guarantee identical bond angles, bond polarities or intermolecular forces. Ammonia has a bond angle of about 107°, while phosphine’s H–P–H angle is much closer to 93.5°, reflecting differences in bonding and orbital participation.
The VSEPR structural comparison is useful context but does not by itself explain the higher boiling point. A student should move from molecular geometry to polarity and then to intermolecular hydrogen bonding, keeping each physical question separate.
An H1 exam answer can usually focus on the polar N–H bond and strong N–H···N intermolecular attraction without requiring a detailed quantum treatment of the phosphine bond angle.
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7. Know the normal boiling-point reference rather than guessing from room temperature
Normal boiling point is the temperature at which a liquid’s equilibrium vapour pressure reaches one atmosphere, or the conventional reference pressure. Ammonia’s normal boiling point around −33.3°C and phosphine’s around −87.7°C mean both substances are gases at ordinary room temperature and one atmosphere.
A larger boiling-point number does not mean the substance must be liquid at room temperature. It means the liquid phase survives to a higher temperature before its vapour pressure reaches the specified external pressure. Under a different pressure, boiling temperatures can change.
Students should not equate boiling with molecular decomposition. A gas can have intact NH₃ or PH₃ molecules after vaporisation. The process involves changes in intermolecular interactions and phase equilibrium rather than the wholesale conversion of molecules into atoms.
The reference values are approximate measured properties compiled in chemical data references, not original measurements made by this tutor site. Always name the pressure context when comparing boiling temperatures.
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8. Explain the usual London dispersion trend without denying the exception
London dispersion forces arise from correlated fluctuations in electron distributions and exist between all neutral molecules, including ammonia and phosphine. Larger and more polarizable electron clouds can provide stronger dispersion forces, contributing to a tendency for boiling temperature to rise within a closely related molecular family.
Phosphine’s greater molecular mass and more polarizable electronic structure could, by dispersion alone, favour a higher boiling point than ammonia. However, ammonia has an additional substantial hydrogen-bonding contribution, overriding that simple mass-based expectation in the measured comparison.
A good explanation must therefore include both effects rather than state that phosphine has zero intermolecular attraction. It is accurate to say dispersion remains important for PH₃ and that NH₃’s hydrogen-bonded liquid is more strongly cohesive than the mass-only trend would predict.
The tutor can teach this as a model-selection habit: list all plausible intermolecular forces, identify the strongest distinctive one, then explain how it affects an observed physical property.
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9. Use arsine as a second control for the mass trend
Arsine, AsH₃, is a heavier group-15 hydride with a normal boiling point around −62.5°C, above phosphine’s approximately −87.7°C. Neither behaves like ammonia’s strong N–H hydrogen-bonding case. The increase from PH₃ to AsH₃ illustrates the familiar effect of stronger dispersion attractions as molecules become heavier and more polarizable.
Ammonia still has a much higher boiling point than both PH₃ and AsH₃ even though it is lighter. Its additional hydrogen bonding explains why the data do not follow one smooth increasing trend from nitrogen downward through group 15.
These comparisons show that using the word anomaly is relative to a particular expected trend. It does not mean the ammonia boiling point violates chemical laws; the initial one-factor model was incomplete.
Arsine and phosphine are highly hazardous substances. This is solely a data-interpretation lesson and contains no instructions to prepare, handle or compare the gases physically.
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10. Why a boiling point is not a direct molecular-bond-energy measurement
A common misconception compares the covalent N–H bond enthalpy with the covalent P–H bond enthalpy and directly predicts the liquid’s boiling point. Bond enthalpy values refer to breaking specified bonds in gaseous molecules, while vaporisation involves transferring intact molecules between phases.
The actual boiling point follows equilibrium thermodynamics, including changes in enthalpy and entropy between liquid and gas. Stronger intermolecular attraction often raises the enthalpy needed for vaporisation, but the final boiling temperature also depends on entropy and pressure.
This level of nuance prevents a correct hydrogen-bonding phrase from becoming an inaccurate universal rule that stronger hydrogen bonds always give one exact predictable temperature. The measured data and structure must be considered together.
Students should not claim a numerical boiling temperature can be calculated solely from electronegativity or relative molecular mass without additional physical data.
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11. Distinguish molecular polarity from a guaranteed hydrogen bond
Ammonia is polar because the N–H bond dipoles and trigonal pyramidal geometry do not cancel. Phosphine can also have a molecular dipole, although the detailed direction and magnitude of its dipole arise from its different electronic structure. Molecular polarity alone, however, does not guarantee strong conventional hydrogen bonding.
Dipole–dipole attraction can contribute to a liquid’s cohesion. Hydrogen bonding is a more specific interaction requiring appropriate donor and acceptor chemistry. If students write ‘NH₃ boils higher because it is polar, whereas PH₃ is nonpolar’ they risk oversimplifying and misrepresenting the phosphine molecule.
The more secure statement is that NH₃ has strongly polar N–H bonds, an N lone pair and intermolecular hydrogen bonding. PH₃ lacks comparable strong conventional hydrogen-bond donor behaviour, so its ordinary intermolecular forces are weaker despite its larger molar mass.
A tutor should teach students to name the actual interaction rather than use polarity as a synonym for every type of cohesion.
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12. Transfer the idea to water and hydrogen fluoride
Water, H₂O, and hydrogen fluoride, HF, also have strongly polar O–H or H–F bonds and can exhibit significant intermolecular hydrogen bonding. Their physical properties reinforce the principle that group-15, -16 or -17 hydride boiling points cannot be predicted from molecular mass alone.
Water has two O–H bonds and two lone pairs on oxygen, making extended hydrogen-bonded networks possible. Hydrogen fluoride has one H–F donor bond and fluorine lone pairs; it also forms strong associated structures. Their precise physical behaviour depends on network patterns and phase conditions.
Students should not rank the three hydrogen-bonding substances from just the count of hydrogen atoms or lone pairs. Boiling temperatures depend on the full thermodynamic behaviour of their liquids, not a universal ‘more hydrogen bonds per molecule equals exact boiling point’ formula.
The comparison is useful as a changed-case question: first establish whether strong hydrogen bonding is possible, then consider size, shape, dispersion and measured physical properties.
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13. Repair the four most common wrong explanations
One mistake says PH₃ must boil higher because it has twice ammonia’s molecular mass. Another claims ammonia boils higher because the intramolecular N–H covalent bond is stronger, without mentioning attraction between different molecules. A third says PH₃ has no intermolecular forces at all. A fourth assumes a lone pair on phosphorus makes PH₃ form exactly the same hydrogen-bond network as NH₃.
The repair is to draw both molecular structures, name the polar N–H donor and neighbouring nitrogen lone-pair acceptor, identify N–H···N attraction, and contrast it with the weaker ordinary PH₃ interactions. The measured normal boiling points then follow the qualitative trend.
Do not invent a claim that a laboratory sample of NH₃ or PH₃ was boiled during this lesson. The chemistry is explained through published thermodynamic data and molecular models.
Parents can ask one question at the end: Why is the heavier molecule easier to separate into gas? A child who explains the absence of strong conventional hydrogen bonding in phosphine has understood the core phenomenon.
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14. Final unseen data task: choose the relevant force first
An original paper task gives four hydrides—NH₃, PH₃, H₂O and CH₄—and asks students to choose the ones able to form significant conventional intermolecular hydrogen bonds. NH₃ and H₂O are the clearest candidates, while PH₃ and CH₄ do not have the same strongly polar N–H or O–H donor arrangement.
The student then compares the measured normal boiling points of NH₃ and PH₃, recognising that hydrogen bonding raises ammonia above the mass-only expectation. The answer should mention that all these molecules still experience dispersion forces.
A second transfer card replaces NH₃ with AsH₃. Without strong hydrogen bonding, the heavier AsH₃ boils above PH₃ in the standard group trend. This shows when a molar-mass or polarizability argument is more predictive.
A complete answer therefore selects the relevant forces before predicting the boiling trend and notes the limitations of the simplified molecular model.
Worked evidence: Group 15 hydrides and useful comparisons
All boiling temperatures are approximate published normal boiling-point reference data, not measurements conducted by eduKateSG. Boiling point refers to a specified ambient-pressure reference, while molecular masses use rounded atomic data.
| Molecule | Approx. Mᵣ | Normal boiling point | Main teaching explanation |
|---|---|---|---|
| NH₃ | 17 | ≈−33.3°C | Strong N–H···N hydrogen bonding |
| PH₃ | 34 | ≈−87.7°C | No comparable strong conventional H bonding |
| AsH₃ | 78 | ≈−62.5°C | Heavier, more polarizable, dispersion trend |
| CH₄ | 16 | ≈−161.5°C | No N–H/O–H/F–H donor; mainly dispersion |
| H₂O | 18 | ≈100°C | Extensive water hydrogen-bond network |
| HF | 20 | ≈19.5°C | Significant intermolecular H bonding |
| NH₃ vs PH₃ | Mass doubles downward | Boiling point falls | Hydrogen bonding overrides mass-only trend |
| PH₃ vs AsH₃ | Mass increases | Boiling point rises | More typical dispersion-dominated trend |
The useful counterexample is the small ammonia molecule: its strong intermolecular hydrogen bonding raises its boiling point well above the more massive PH₃. In the PH₃/AsH₃ comparison, dispersion and polarizability provide a more familiar increasing-mass trend.
Four independent diagnostic tasks
- Compare NH₃ and PH₃, reporting Mᵣ 17 versus 34 and boiling points about −33.3 versus −87.7°C, then explain the apparent anomaly.
- Draw N–H···N between two ammonia molecules and label covalent N–H bonds separately from the intermolecular attraction.
- Compare PH₃ with AsH₃ and explain why the heavier arsine has a higher boiling point without claiming strong N–H-like hydrogen bonding.
- Given NH₃, PH₃, CH₄ and H₂O, identify plausible strong conventional hydrogen-bond donors and explain why a lone pair alone is insufficient.
Tutor checklist
- Confirm the 2027 H1 Chemistry 8873 Chemical Bonding and physical-property outcomes.
- Calculate molecular masses without inferring boiling point from mass alone.
- Draw trigonal pyramidal NH₃ and PH₃ structures with lone pairs.
- Identify polar N–H donors and nitrogen lone-pair acceptors.
- Distinguish intermolecular hydrogen bonding from intramolecular covalent bonds.
- Recognise London dispersion forces in all neutral molecules.
- Use normal boiling-point values with their reference pressure and negative Celsius signs.
- Compare the PH₃/AsH₃ dispersion trend and NH₃ exception.
- Keep high-toxicity phosphine and arsine discussion theoretical.
- Finish with an unfamiliar hydride and a qualified structural explanation.
Frequently asked questions
Why does ammonia boil above phosphine?
Ammonia molecules can form significant intermolecular N–H···N hydrogen bonds, unlike the weaker ordinary interactions in phosphine.
Is phosphine heavier than ammonia?
Yes. Approximate relative molecular masses are PH₃ 34 and NH₃ 17.
Are the normal boiling points positive temperatures?
No. Ammonia is about −33.3°C and phosphine about −87.7°C at ordinary reference pressure.
Does PH₃ lack all intermolecular forces?
No. It experiences London dispersion and other relatively weak interactions.
Does a P lone pair make PH₃ a strong hydrogen-bonded liquid?
Not by itself. The P–H donor bond is not sufficiently polar for the strong conventional N–H-like pattern.
Are N–H covalent bonds broken when ammonia boils?
No. Boiling separates intact molecules and changes their intermolecular environment.
Why does arsine boil higher than phosphine?
Greater molecular mass and polarizability contribute to stronger dispersion interactions in that comparison.
Is ammonia’s higher boiling point predicted by molar mass alone?
No. The mass-only trend misses its additional hydrogen bonding.
Does the 2027 H1 syllabus cover this?
Yes. The chemical-bonding and intermolecular-force concepts support explanation of such physical-property comparisons.
Connected bonding guides
- JC1 H1 Chemistry | Ionic Lattice versus Metallic Deformation
- JC1 H1 Chemistry | Neutral Water and Temperature
- Hougang Chemistry Tuition and Tutor Library
Bring the learner’s exam year and a chemical-bonding comparison question to discuss tutorial support. Confirm actual class availability, teacher, fees, location and schedule directly. No outcome or vacancy is guaranteed.
Official curriculum and measured physical-property sources
- SEAB 2027 A-Level syllabuses for school candidates, including H1 Chemistry 8873
- MOE Pre-University Chemistry 8873/9476/9813 curriculum syllabus
- NIST Chemistry WebBook: Phosphine Phase-Change Properties
- PubChem: Phosphine Normal Boiling Point and Safety Context
- Immutable eduKateSG Clementi small-group teaching reference
- Hougang Chemistry Tuition and Tutor Library
Hydrogen bonding and dispersion are explained as qualitative structure-property models. Approximate normal boiling points are published reference data, and the comparison does not constitute an experiment. Phosphine and arsine are toxic gases; this article includes no handling or production instructions.