Hydrogen bonding & properties
Hydrogen bonding & properties
Hydrogen bonding: conditions and diagrams
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Hydrogen bonding needs a particular arrangement. Find a δ+ hydrogen directly bonded to N, O or F, then a lone pair on N, O or F in a neighbouring molecule.
- Find a δ+ H atom directly bonded to N, O or F in one molecule.
- Find a lone pair on N, O or F in a neighbouring molecule.
- Draw a dotted attraction from that lone pair to the δ+ H. The covalent bond from H to its own atom stays in place.
- Add the appropriate partial charges and keep the hydrogen-bond arrangement clear.
Water, ammonia and HF can hydrogen-bond between their own molecules. An oxygen lone pair can attract a δ+ H atom in water, even when the oxygen-containing molecule has no O-H bond and cannot hydrogen-bond with itself. H bonded only to carbon does not meet the hydrogen-bonding condition here.
- Hydrogen bond
- An attraction between a lone pair on N, O or F and a δ+ H atom covalently bonded to N, O or F in a neighbouring molecule.
| Substance | Self hydrogen-bonding? | Reason |
|---|---|---|
| HF | Yes | H–F bonds and F lone pairs |
| H2O | Yes | O–H bonds and O lone pairs |
| NH3 | Yes | N–H bonds and an N lone pair |
| CH3F | No | H is bonded to C, not F |
| CH3OCH3 | No | O lone pairs but no H directly bonded to O |
Propan-1-ol and methoxyethane
Propan-1-ol and methoxyethane have the same molecular formula. Explain why propan-1-ol has the higher boiling point.
Worked answer
Both have van der Waals and permanent dipole–dipole forces. Propan-1-ol also self hydrogen-bonds through O–H groups; methoxyethane has no H directly bonded to N, O or F.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Draw a hydrogen bond between two water molecules, showing relevant partial charges and a lone pair on oxygen. [4]
- Explain why HCl molecules do not hydrogen-bond to one another. [1]
- State why a hydrogen bond must not be drawn as a solid covalent bond. [1]
Total: 6 marks
Show answer & marking points
- O–H covalent bond [1]; Hδ+ and Oδ− on the neighbouring molecule [1]; oxygen lone pair [1]; dotted attraction from Hδ+ to that lone pair [1].
- H is not directly bonded to N, O or F [1].
- It is an intermolecular electrostatic attraction, not a shared pair of electrons [1].
| Hydride | H2O | H2S | H2Se | H2Te |
|---|---|---|---|---|
| Boiling point / °C | 100 | −60 | −41 | −2 |
- Explain the trend from H2S to H2Te. [3]
- Explain water’s anomalously high boiling point. [3]
- Draw one hydrogen bond between water molecules, with partial charges and an oxygen lone pair. [4]
Total: 10 marks
Show answer & marking points
- Electron count and polarisability increase [1]; van der Waals forces strengthen [1], so more energy is required [1].
- Water forms hydrogen bonds [1]; its overall attractions are stronger than those in the other hydrides [1], so more energy is needed [1].
- Hδ+ in an O–H bond [1]; Oδ− on the neighbouring molecule [1]; O lone pair [1]; dotted attraction from H to the lone pair [1].
Hydrogen bonding, ice and anomalous boiling points
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Hydrogen bonds help explain two unusual properties of water. They make water's boiling point higher than the simple trend for comparable hydrides suggests, and they give ice its open structure.
In ice, hydrogen bonds hold water molecules further apart than they are on average in liquid water. The same mass occupies a larger volume, so ice is less dense and floats.
On melting, some hydrogen bonds are overcome and the open arrangement partly collapses. The O-H covalent bonds inside each water molecule remain intact.
Use the same link for boiling-point comparisons: stronger overall attractions between molecules mean more energy is needed to separate them.
| Comparison | Explanation |
|---|---|
| NH3 versus PH3 | NH3 forms hydrogen bonds; its boiling point is unusually high |
| HF versus HCl | HF forms hydrogen bonds; more energy is required to separate molecules |
| H2O versus other Group 6 hydrides | Water’s hydrogen-bonded network raises its boiling point |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
The figure shows a simplified open structure of ice.
- Identify the dotted attraction and the solid bond. [2]
- Explain why ice is less dense than liquid water. [3]
- State what is overcome on melting and what remains unbroken within each molecule. [2]
Total: 7 marks
Show answer & marking points
- Dotted: hydrogen bonds [1]; solid O–H: covalent bonds [1].
- Hydrogen bonds hold an open lattice [1]; molecules are further apart and the same mass occupies a larger volume [1], so density is lower [1].
- Some intermolecular hydrogen bonds are overcome as the lattice collapses [1]; O–H covalent bonds remain intact [1].
Boiling, melting and competing intermolecular effects
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To compare boiling points, explain what has to be overcome when the particles separate. Stronger overall intermolecular attractions need more energy, so boiling point tends to be higher.
- Identify the molecules and all the forces between them.
- Compare electron count, polarisability, shape and any extra permanent-dipole or hydrogen-bonding attractions.
- Decide which substance has stronger overall attractions.
- Link that to more energy needed and the higher boiling point.
A more volatile substance evaporates more readily. Stronger attractions generally reduce volatility at a given temperature. Melting points also depend on how well particles pack in the crystal, so avoid using one simple force ranking for every comparison.
| Process | Overcome | Remains intact |
|---|---|---|
| Boiling water | Intermolecular hydrogen bonds | O–H covalent bonds |
| Melting iodine | Intermolecular van der Waals forces | I–I covalent bonds |
Ethanol and ethane
Explain why ethanol has a higher boiling point than ethane.
Worked answer
Both have van der Waals forces. Ethanol is polar and also forms hydrogen bonds. Its overall attractions are stronger, so more energy is needed to separate molecules.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
| Substance | Formula | Mᵣ | Boiling point / °C |
|---|---|---|---|
| A | CH3CH2CH3 | 44 | −42 |
| B | CH3CHO | 44 | 21 |
| C | CH3CH2OH | 46 | 78 |
| D | HOCH2CH2OH | 62 | 197 |
- State the strongest intermolecular force in each. [4]
- Explain why B boils above A despite equal Mᵣ. [2]
- Explain why C boils above B. [2]
- Explain why D boils above C. [3]
- Place A–D in increasing volatility. [1]
Total: 12 marks
Show answer & marking points
- A van der Waals; B permanent dipole–dipole; C hydrogen bonding; D hydrogen bonding. One mark each.
- B has permanent dipole–dipole attractions as well as van der Waals forces [1], giving stronger overall attractions and greater energy needed than A [1].
- C forms hydrogen bonds [1], so its overall attractions require more energy to overcome than B’s [1].
- D has two O–H groups and a more extensive hydrogen-bonded network [1], plus more electrons and stronger van der Waals forces [1], so greater energy is needed [1].
- D < C < B < A [1].
Three liquid jets are tested with the same charged rod.
| Jet | Deflection | Mᵣ | Boiling point / °C |
|---|---|---|---|
| P | Strong | 18 | 100 |
| Q | Moderate | 58 | 56 |
| R | Negligible | 86 | 69 |
- Identify P, Q and R from water, propanone and hexane. [3]
- Explain why a charged rod deflects a polar liquid jet. [2]
- Explain why deflection alone does not prove self hydrogen bonding. [2]
- Explain why P boils highest despite having the lowest Mᵣ. [2]
- Explain why non-polar R boils above Q. [3]
Total: 12 marks
Show answer & marking points
- P water [1]; Q propanone [1]; R hexane [1].
- The electric field aligns / attracts permanent molecular dipoles [1], giving a net sideways force [1].
- In this comparison, deflection indicates polarity rather than uniquely identifying hydrogen bonding [1]. Propanone deflects but has no H directly bonded to N, O or F, so it cannot hydrogen-bond with itself [1].
- Water forms hydrogen bonds [1], requiring more energy to overcome its overall attractions [1].
- Hexane has more electrons and greater polarisability [1]; its stronger van der Waals forces [1] outweigh propanone’s permanent dipole contribution [1].
Solubility: attractions disrupted and formed
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Dissolving means particles of the solute mix with the solvent. Some attractions within the solute and between solvent particles are disrupted, while new attractions form between solute and solvent.
Name those new attractions. Ethanol can hydrogen-bond with water. Propanone can accept hydrogen bonds from water through its oxygen lone pairs, even though propanone does not hydrogen-bond to itself.
Hexane cannot form hydrogen bonds that adequately replace water-water attractions, so its solubility in water is very low. Iodine mixes more readily with non-polar hexane through favourable van der Waals interactions.
'Like dissolves like' is a useful starting point. An exam explanation needs the actual particles and attractions, rather than the slogan alone.
| Pair | Interaction reasoning | Prediction |
|---|---|---|
| Ethanol + water | O–H groups and O lone pairs allow hydrogen bonds between the two substances | Mix readily |
| Hexane + water | Hexane cannot form hydrogen bonds that adequately replace water–water attractions | Very low solubility |
| Iodine + hexane | Both are non-polar; favourable van der Waals interactions form | Iodine dissolves better than in water |
| Propanone + water | O lone pairs attract δ+ H atoms in water | Mix readily |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Ethanol, CH3CH2OH, boils at 78 °C. Methoxymethane, CH3OCH3, boils at −25 °C. Both have formula C2H6O.
- State the strongest intermolecular force in each. [2]
- Explain the boiling-point difference. [3]
- Both hydrogen-bond with water. Explain and identify one difference in their hydrogen-bonding behaviour. [2]
Total: 7 marks
Show answer & marking points
- Ethanol: hydrogen bonding [1]; methoxymethane: permanent dipole–dipole [1].
- Ethanol’s O–H groups allow self hydrogen bonding [1]. Methoxymethane has no H bonded to N, O or F, so it cannot self hydrogen-bond [1]. Stronger overall attractions in ethanol require more energy to overcome [1].
- Both have O lone pairs that attract δ+ H atoms in water [1]. Ethanol also has a δ+ H in an O–H bond that can hydrogen-bond to a water oxygen lone pair; methoxymethane has no O–H bond [1].