Hydrogen bonding & properties
Hydrogen bonding & properties
Hydrogen bonding and correct 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 intermolecular attraction between a lone pair on N, O or F in one molecule and a δ+ H atom covalently bonded to N, O or F in another molecule.
| Check | Requirement |
|---|---|
| Hydrogen atom | H directly bonded to N, O or F. |
| Lone pair | An available lone pair on N, O or F. |
| Drawing | Solid covalent bonds; dotted intermolecular line from the neighbouring N/O/F lone pair to δ+ H; correct δ signs. |
Water compared with hydrogen sulfide
Why does H2O have a much higher boiling point than H2S?
Worked answer
- H2O has O–H bonds and oxygen lone pairs, so its molecules hydrogen-bond.
- H2S does not form the same hydrogen-bonding network.
- More energy is needed to overcome the attractions between water molecules.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) Draw two water molecules and show one hydrogen bond. Include δ+ and δ-, a lone pair, solid covalent bonds and a dotted intermolecular attraction. [5 marks]
(b) Identify the two atoms linked by the hydrogen bond in your diagram, and explain why the dotted line is not an O-H covalent bond. [2 marks]
Total: 7 marks
Show answer & marking points
(a) Two correct bent H2O molecules with solid O-H bonds [1]; O δ- and H δ+ [1]; a lone pair on O [1]; dotted H···O attraction [1]; line joins δ+ H to a lone pair / δ- O of the other molecule [1]. [5]
(b) The hydrogen bond joins the δ+ H of one water molecule to the lone pair on O of another [1]. The dotted line is an intermolecular electrostatic attraction, not a shared electron pair / covalent bond [1]. [2]
The boiling points of four hydrogen halides are shown.
| Substance | Boiling point / °C |
|---|---|
| HF | 19.5 |
| HCl | −85 |
| HBr | −67 |
| HI | −35 |
- Explain why HF has a much higher boiling point than HCl. [4]
- Explain the increase in boiling point from HCl to HI. [4]
Total: 8 marks
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(a) F is highly electronegative, so HF has a strongly polar H-F bond [1]. The H δ+ of one molecule is attracted to a lone pair on F of another [1], forming hydrogen bonds [1]. More energy is required to overcome these stronger attractions [1]. [4]
(b) HCl to HI have progressively more electrons / larger electron clouds [1], which are more easily polarised [1]. Induced dipoles and London forces therefore become stronger [1], so more energy is needed and boiling point rises [1]. [4]
Ethanol, CH3CH2OH, and methoxymethane, CH3OCH3, have the same molecular formula and relative molecular mass. Their boiling points are 78.4 °C and -24.8 °C respectively.
(a) State the strongest intermolecular force in each pure liquid. [2 marks]
(b) Explain the large difference in boiling point. [4 marks]
(c) Draw one hydrogen bond between ethanol and water. Show a lone pair and correct partial charges. [2 marks]
Total: 8 marks
Show answer & marking points
(a) Ethanol: hydrogen bonding [1]. Methoxymethane: permanent dipole-dipole attraction [1]. [2]
(b) Ethanol contains an O-H bond and can form hydrogen bonds between its molecules [1]. Methoxymethane has no H directly bonded to O, N or F and cannot hydrogen-bond with itself [1]. Hydrogen bonds are stronger than its permanent dipole-dipole attractions [1], so more energy is required to separate ethanol molecules [1]. [4]
(c) Correct O-H bond, lone pair on the neighbouring O, δ signs and dotted H···O attraction [2]. [2]
Hydrazine has the structural formula H2N-NH2. Each nitrogen atom has three bonded pairs and one lone pair. Hydrazine boils at 114 °C; ethane, C2H6, boils at -89 °C. Their relative molecular masses are similar.
(a) State the shape name around each nitrogen atom and state how its bond angle compares with 109.5°. A drawing is not required in part (a). [3 marks]
(b) Draw one hydrogen bond between two hydrazine molecules. Show the N-H bond, the neighbouring nitrogen lone pair and partial charges. [2 marks]
(c) Explain the large difference between the boiling points of hydrazine and ethane. [2 marks]
Total: 7 marks
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(a) Each N has four electron pairs, one of which is lone, so the shape around each N is pyramidal [2]. Lone-pair repulsion compresses the angle below 109.5° [1]. [3]
(b) Dotted N-H···N attraction from δ+ H in an N-H bond [1] to a lone pair on the neighbouring δ− N [1]. [2]
(c) Hydrazine forms hydrogen bonds whereas ethane has only induced dipole-dipole attractions [1]; more energy is needed to separate hydrazine molecules [1]. [2]
A student’s proposed hydrogen-bond diagram for two ethanol molecules is shown.
- Identify three errors in the diagram. [3]
- Redraw the interaction correctly. [3]
Total: 6 marks
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(a) O atoms should be δ-, not δ+ [1]; H in O-H should be δ+, not δ- [1]; the intermolecular attraction must be dotted / broken rather than drawn as a solid covalent-bond line [1]. [3]
(b) Correct solid covalent bonds [1], δ+ H···O δ- dotted attraction [1] and a lone pair on oxygen in the neighbouring molecule [1]. [3]
Water’s boiling point and the density of ice
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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.
| Anomaly | Explanation |
|---|---|
| Relatively high melting and boiling points | Many hydrogen bonds require a substantial energy input to overcome enough attractions for molecules to move or separate. |
| Ice is less dense than liquid water | An ordered open lattice holds water molecules farther apart. Part of the network collapses on melting. |
Why does ice float?
Explain why solid ice is less dense than liquid water and therefore floats on water.
Worked answer
- Hydrogen bonds hold water molecules in an open, ordered lattice.
- For the same mass, ice occupies a larger volume.
- Density = mass ÷ volume, so ice is less dense.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) Draw part of the open hydrogen-bonded lattice in ice. Distinguish O-H covalent bonds from hydrogen bonds. [3 marks]
(b) Explain why ice is less dense than liquid water. [3 marks]
(c) Explain why water has relatively high melting and boiling points. [2 marks]
Total: 8 marks
Show answer & marking points
(a) Water molecules are linked in an open arrangement by dotted hydrogen bonds [1], with solid O-H covalent bonds clearly distinguished [1] and the dotted attraction joining H δ+ to O δ- / a lone pair [1]. [3]
(b) In ice, hydrogen bonds hold water molecules in a fixed open lattice [1] with relatively large spaces [1]. The same mass therefore occupies a larger volume than in liquid water, so ice has lower density [1]. [3]
(c) Many hydrogen bonds between water molecules must be overcome [1], requiring relatively large energy input [1]. [2]
Molecular lattices, solubility and volatility
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For a molecular substance, separate the strong covalent bonds inside the molecules from the attractions between them. Melting and boiling normally overcome the between-molecule attractions.
Stronger overall attractions tend to increase boiling point and reduce volatility. Neutral molecules normally provide no mobile ions or delocalised electrons, so the pure molecular substance does not conduct.
For solubility, identify the attractions disrupted and the new ones formed with the solvent. A molecule can accept a hydrogen bond from water without being able to hydrogen-bond to itself. The examples below show how to name the relevant interactions.
- Simple molecular lattice
- Discrete covalently bonded molecules held together in a solid by intermolecular forces.
| Property | Explanation |
|---|---|
| Melting / boiling | Intermolecular forces are overcome; molecules usually stay intact. |
| Pure electrical conductivity | No mobile ions or delocalised electrons. |
| Solubility | Depends on new solute-solvent attractions compared with those disrupted. |
| Volatility | Weaker attractions usually mean a lower boiling point and higher volatility. |
Iodine versus diamond
Explain why iodine has a much lower melting point than diamond. Refer to their structures and the attractions that must be overcome.
Worked answer
Iodine is simple molecular, so melting overcomes attractions between I2 molecules. Diamond is a giant covalent network whose strong C–C bonds would need to be broken.
Iodine in cyclohexane
Why is iodine much more soluble in cyclohexane than in water?
Worked answer
I2 and cyclohexane are non-polar. Favourable induced dipole-dipole attractions can form between them, whereas iodine does not replace water’s hydrogen-bond attractions as effectively.
Methanol in water
Why does methanol mix completely with water?
Worked answer
Methanol has an O–H group and oxygen lone pairs. Methanol and water form favourable hydrogen bonds with one another.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Consider iodine, I2, methanol, CH3OH, and hexane, C6H14.
(a) Predict whether iodine is more soluble in water or in hexane. Explain your answer using intermolecular forces. [3 marks]
(b) Predict whether methanol is more soluble in water or in hexane. Explain your answer. [3 marks]
(c) Explain why pure liquid methanol does not conduct electricity. [1 mark]
Total: 7 marks
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(a) I2 is more soluble in hexane [1]. Both are non-polar and interact mainly through induced dipole-dipole forces [1]; comparable new attractions can replace those broken on mixing [1]. [3]
(b) Methanol is more soluble in water [1]. Both are polar and methanol can form hydrogen bonds with water [1], giving favourable solute-solvent attractions [1]. [3]
(c) Pure methanol contains neutral molecules and has no mobile ions or delocalised electrons to carry charge [1]. [1]
Build complete intermolecular-force comparisons
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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.
| Weak explanation | Improve it |
|---|---|
| “NH3 has stronger bonds” | Name hydrogen bonds between molecules, not N–H covalent bonds. |
| “I2 is bigger” | Explain electron count, polarisability, induced dipoles, attraction and energy. |
| “CO2 bonds are non-polar” | The C=O bonds are polar; equal opposite dipoles cancel. |
Ammonia compared with phosphine
Why does NH3 have a higher boiling point than PH3?
Worked answer
NH3 contains N–H bonds and a nitrogen lone pair, so its molecules hydrogen-bond. PH3 does not. More energy is needed to separate NH3 molecules.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Each carbon atom in the following chloromethanes is approximately tetrahedral.
| Molecule | Boiling point / °C |
|---|---|
| CH3Cl | −24 |
| CH2Cl2 | 40 |
| CHCl3 | 61 |
| CCl4 | 77 |
- Explain why every C–Cl bond is polar. [2]
- Determine which molecules are polar. For the non-polar molecule, name its shape and explain dipole cancellation. A drawing is not required. [4]
- Explain why CCl4 can have the highest boiling point despite being non-polar. [2]
Total: 8 marks
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(a) Cl is more electronegative than C [1], so electron density is drawn towards Cl and each C-Cl bond has C δ+ and Cl δ- [1]. [2]
(b) CH3Cl, CH2Cl2 and CHCl3 are polar [3]. CCl4 is tetrahedral and symmetrical, so its four equal C-Cl dipoles cancel and it is non-polar [1]. [4]
(c) CCl4 has the largest electron cloud / most electrons [1], so it has the strongest induced dipole-dipole forces and needs the most energy to separate its molecules [1]. [2]
Three substances have similar relative molecular masses but very different boiling points.
| Substance | Formula | Mᵣ | Boiling point / °C |
|---|---|---|---|
| Ethane | C2H6 | 30 | −89 |
| Methanal | CH2O | 30 | −19 |
| Methanol | CH3OH | 32 | 65 |
- State the strongest intermolecular force in each substance. [3]
- Explain the boiling-point order, distinguishing all three types of intermolecular force. [4]
Total: 7 marks
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(a) Ethane: induced dipole-dipole [1]. Methanal: permanent dipole-dipole [1]. Methanol: hydrogen bonding [1]. [3]
(b) Similar Mr values reduce the difference in London forces [1]. Methanal has a permanent C=O dipole, giving stronger attraction than in non-polar ethane [1]. Methanol molecules form hydrogen bonds [1], the strongest interaction in the set, so methanol requires the most energy to boil [1]. [4]
Combine shape, polarity and physical properties
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A longer question may connect shape, polarity and boiling point. Work through those ideas in that order so one answer supports the next.
- Count the central-atom electron regions, including lone pairs.
- Draw and name the three-dimensional molecular shape.
- Use that shape to decide whether the bond dipoles cancel.
- Identify all the applicable intermolecular forces, including induced dipoles.
- Explain the property using energy needed to separate particles, or the mobility of charge carriers.
Keep the reasoning visible. A correct shape or a named force on its own is only part of the explanation.
BF3 has polar B–F bonds
BF₃ contains polar B–F bonds. Predict its molecular shape, bond angle and overall polarity, explaining your reasoning.
Worked answer
- Boron has three bonding regions and no lone pairs.
- The regions arrange trigonal planar at 120°.
- Three equal B–F dipoles are symmetrically arranged and cancel.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Molecules X, Y and Z are CH3F, CF4 and CH3OH respectively. Their boiling points are -78 °C, -128 °C and 65 °C. All three have four electron regions around carbon.
(a) For both CH3F and CF4, state the shape name and bond angle around carbon, then draw a clear 3-D diagram. [3 marks]
(b) Explain why CH3F is polar but CF4 is non-polar. [3 marks]
(c) State the strongest intermolecular force in each of X, Y and Z. [2 marks]
(d) Explain the boiling-point order Y < X < Z. [2 marks]
Total: 10 marks
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(a) Both molecules are tetrahedral around carbon with a bond angle of approximately 109.5° [1]. Award one mark for each of two clear 3-D wedge-and-dash diagrams with four bonds and similar bond lengths [2]. [3]
(b) In CH3F, the polar C-F bond produces a resultant dipole that is not cancelled by the three C-H bond dipoles [2]. In CF4, four identical C-F dipoles are arranged symmetrically and cancel [1]. [3]
(c) X CH3F: permanent dipole-dipole; Y CF4: induced dipole-dipole; Z CH3OH: hydrogen bonding. Award [2] for all three correct or [1] for any two. [2]
(d) CF4 has only induced dipole forces; CH3F also has permanent dipole attractions; CH3OH forms hydrogen bonds, so progressively more energy is needed to separate the molecules [2]. [2]