Module 1 · L10 of 2140 min⚡ +50 XP in Learn · +25 to completeYear 11 · Module 1 · IQ4
Intermolecular Forces and Physical Properties
Today's hook, Water boils at 100°C. Hydrogen sulfide is a heavier molecule, and it boils at about minus 60°C. By size alone, water should be a gas in this room.
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Warm up and recall
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Water (H₂O) and hydrogen sulfide (H₂S) have similar molecular shapes, yet water is a liquid at room temperature with a boiling point of 100°C, while H₂S is a gas with a boiling point of −60°C. What could explain such a huge difference in boiling point between two substances with such similar formulas?
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What you'll master, and the words for it
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What you'll master
Know
Key facts
The three types of IMFs: dispersion, dipole-dipole, and hydrogen bonding
Which molecular features give rise to each type of IMF
How IMF type and strength determine physical properties
Understand
Concepts
Why dispersion forces increase with molecular size and shape
Why polarity and electronegativity determine which IMFs are present
How to predict and explain BP trends within a homologous series
Can do
Skills
Identify which IMFs are present in a given molecule
Rank molecules by predicted BP and justify the ranking
Analyse data tables to identify IMF patterns and explain anomalies
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The Three Types of Intermolecular Forces
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The Three Types of Intermolecular Forces
core concept
Dispersion forces
Present in:ALL molecules (polar and non-polar)
Relative strength:Weakest (increases with size)
Cause:Instantaneous temporary dipoles from electron cloud fluctuations
Example molecules:CH₄, CO₂, I₂, noble gases, hydrocarbons
Dipole-dipole forces
Present in:Polar molecules only (non-polar molecules excluded)
Relative strength:Moderate (in addition to dispersion)
Cause:Permanent partial charges (δ+/δ−) due to electronegativity differences in bonds
Example molecules:HCl, SO₂, CHCl₃, acetone (CH₃COCH₃)
Hydrogen bonding
Present in:Molecules with N–H, O–H, or F–H bonds
Relative strength:Strongest IMF per interaction (~5–10× one dipole-dipole contact)
Cause:Highly polar H···X bond (X = F, O, N); lone pair on adjacent F, O, or N
Example molecules:H₂O, HF, NH₃, alcohols, DNA base pairs
Important rule: Dispersion forces are present in ALL molecules, they do not disappear just because a molecule also has dipole-dipole or hydrogen bonding. In polar molecules, both dispersion AND dipole-dipole act. In molecules with N-H/O-H/F-H, ALL THREE types may act. When ranking IMF strength for large polar molecules, dispersion forces can dominate over dipole-dipole if the molecule is large enough. Hydrogen bonding is the strongest force per contact, but total attraction is the sum of every contact: a large, highly polarisable molecule with only dispersion forces can boil higher than a small hydrogen-bonding molecule, so never rank by the strongest label alone, weigh how many interactions act and the molecule's size and shape too.
Three intermolecular forces (IMFs), weakest → strongest: dispersion (present in ALL molecules, from temporary dipoles in fluctuating electron clouds) < dipole–dipole (polar molecules only) < hydrogen bonding (only when H is bonded to N, O, or F, and an adjacent N/O/F lone pair accepts). Higher total IMF strength → higher BP.
Pause, copy the highlighted IMF ranking into your book before moving on.
Quick check: which list correctly orders the three intermolecular forces from weakest to strongest?
Dispersion Forces in Detail
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Dispersion Forces in Detail
core concept
How do dispersion forces arise?
Even in non-polar molecules, electrons are constantly moving. At any instant, the electron distribution may be asymmetric, creating a temporary (instantaneous) dipole. This temporary dipole induces a dipole in the adjacent molecule. The result is a weak, fleeting attraction.
What determines the strength of dispersion forces?
Number of electrons: More electrons → larger, more polarisable electron cloud → stronger temporary dipoles → stronger dispersion forces. This is why BP increases down a group of non-polar molecules (e.g. noble gases, halogens, alkanes).
Molecular shape: Long, straight-chain molecules can contact each other along their entire length → stronger dispersion forces than branched isomers of the same formula. e.g. n-pentane (BP 36°C) > neopentane (BP 9.5°C) despite identical molecular formulae (C₅H₁₂).
Homologous series BP trend: In a series like CH₄ → C₂H₆ → C₃H₈ → C₄H₁₀ (alkanes), each addition of,CH₂, adds ~20 electrons → stronger dispersion forces → higher BP. This creates a smooth, predictable increase in BP with carbon chain length.
We just saw the three IMF types and their relative strengths. That raises a question: what exactly causes dispersion forces and why do larger molecules have stronger ones? This card answers it → dispersion forces arise from temporary electron-cloud fluctuations; more electrons means a more polarisable cloud and thus stronger forces.
Dispersion forces arise from instantaneous dipoles in fluctuating electron clouds. More electrons → more polarisable cloud → stronger dispersion (e.g. I₂ > Br₂ > Cl₂ > F₂). Molecular shape matters: straight chains contact along full length → stronger dispersion than branched isomers (n-pentane BP 36°C vs neopentane 9.5°C). In a homologous series, each extra,CH₂, raises BP by ~20°C.
Add the highlighted dispersion force rule to your notes before the check below.
Match: connect each non-polar molecule to the property that best explains its boiling point.
CH₄ (BP −162°C)
I₂ (BP 184°C)
n-pentane (BP 36°C)
neopentane (BP 9.5°C)
Same formula as its branched isomer but straight chain, molecules contact along full length, stronger dispersion.
Many electrons in a large, very polarisable cloud → strong dispersion forces → high BP.
Compact, branched shape limits contact area between molecules → weaker dispersion than its straight-chain isomer.
Only 10 electrons → very weak dispersion forces → very low BP.
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Hydrogen Bonding in Detail
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Hydrogen Bonding in Detail
core concept
Requirements for hydrogen bonding
Two conditions must BOTH be met:
Hydrogen must be covalently bonded to F, O, or N these are the only atoms electronegative enough to create the large partial positive charge on H needed for H-bonding.
An adjacent molecule must have a lone pair on F, O, or N this lone pair acts as the acceptor.
Common misconception: H–Cl is polar, but HCl does NOT form hydrogen bonds. Chlorine is not electronegative enough (χ = 3.0 vs F = 4.0, O = 3.5, N = 3.0, Cl is borderline but generally not considered). HCl has only dispersion + dipole-dipole forces → lower BP than expected for hydrogen bonding.
Consequences of hydrogen bonding
H₂O
IMFs present:Dispersion + H-bonding (O–H···O)
BP (°C):100
Expected BP without H-bonding:~−80 (extrapolating H₂S trend)
Anomaly:+180°C anomaly
HF
IMFs present:Dispersion + H-bonding (F–H···F)
BP (°C):19.5
Expected BP without H-bonding:~−100 (extrapolating HCl trend)
Anomaly:+120°C anomaly
NH₃
IMFs present:Dispersion + H-bonding (N–H···N)
BP (°C):−33
Expected BP without H-bonding:~−100 (extrapolating PH₃ trend)
Anomaly:+67°C anomaly
H₂S
IMFs present:Dispersion + dipole-dipole only
BP (°C):−60
Expected BP without H-bonding:
Anomaly:No anomaly (no H-bonding)
Water's unique properties from H-bonding: Not just BP, hydrogen bonding also explains water's high surface tension, high heat capacity, and the fact that ice is less dense than liquid water (H-bonds hold molecules in an open hexagonal lattice in ice, farther apart than in liquid water).
We just saw that dispersion forces depend on electron count and molecular shape. That raises a question: why do water, HF, and NH₃ have much higher BPs than expected for their molecular masses? This card answers it → hydrogen bonding, which requires very specific structural conditions, adds an extra layer of intermolecular attraction.
Hydrogen bonding requirements: (1) H must be covalently bonded to N, O, or F; (2) an adjacent molecule must have a lone pair on N, O, or F. HCl does not H-bond (Cl not electronegative enough). H-bonding explains water's anomalously high BP (100°C vs H₂S −60°C), high surface tension, high heat capacity, and why ice is less dense than liquid water.
Pause, write the highlighted H-bonding requirements into your book.
Write it out: in one or two sentences, explain why HCl does NOT form hydrogen bonds even though the H–Cl bond is polar.
Model answer: Hydrogen bonding only forms when H is bonded to F, O or N, atoms with high enough electronegativity to create the large partial positive charge on H needed for the interaction. Chlorine is less electronegative than F/O/N, so H–Cl is polar but not polar enough for true hydrogen bonding.
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Predicting and Comparing Boiling Points
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Predicting and Comparing Boiling Points
core concept
Step-by-step approach
When comparing molecules of very different sizes: Even if one molecule has hydrogen bonding, a much larger non-polar molecule may have a higher BP because its very strong dispersion forces dominate. Always consider both IMF type AND molecular size.
We just saw when and why hydrogen bonding occurs. That raises a question: in an exam question asking you to rank or compare BPs of several substances, what systematic approach guarantees you don't miss any IMF? This card answers it → follow the four-step flowchart: H-bonding check → polarity → size → rank.
BP-prediction flowchart: Step 1, H bonded to N/O/F? → H-bonding. Step 2, polar molecule? → dipole–dipole. Step 3, compare electron count and shape → dispersion strength. Higher total IMF → higher BP. Note: a very large non-polar molecule can outrank a small H-bonded one (size effect can dominate).
Add the highlighted prediction flowchart to your notes before the check below.
Quick check: using the flowchart, which is the FIRST question to ask when predicting the strongest IMF?
Short Answer Questions
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Short Answer Questions
core concept
6. Explain why the boiling point of water (100°C) is much higher than that of hydrogen sulfide (H₂S, −60°C), even though H₂S is a larger, heavier molecule. In your answer, identify all the IMFs present in each substance and explain which is stronger. 4 MARKS
✏️ Answer in your book
7. A student is given three substances with BP data: fluoromethane (CH₃F, BP −78°C), methanol (CH₃OH, BP 65°C), and methane (CH₄, BP −162°C). (a) Identify the IMFs present in each substance. (b) Explain the BP trend in terms of IMF strength. 5 MARKS
✏️ Answer in your book
8. Evaluate the statement: "The boiling point of a substance depends entirely on the type of intermolecular force present, a substance with hydrogen bonding will always have a higher boiling point than one with only dispersion forces." Is this statement correct? Provide evidence and reasoning to support your answer. 4 MARKS
✏️ Answer in your book
We just saw the BP-prediction flowchart. That raises a question: what exact structure do examiners expect when you compare BPs in a written answer? This card answers it → name all IMFs, identify the strongest, link it to higher energy needed to separate molecules, conclude BP ranking.
For BP-comparison exam answers: name ALL IMFs present in each substance (dispersion is in every molecule), then identify the strongest and explain why it is strongest. Example: H₂O > H₂S because water H-bonds (O–H···O) while H₂S has only dispersion + dipole–dipole. Always mention "more energy is required to overcome the stronger IMFs".
Pause, copy the highlighted answer structure into your book before moving on.
Match: connect each substance to the strongest IMF acting between its molecules.
CH₄ (methane)
CH₃F (fluoromethane)
CH₃OH (methanol)
HF (hydrogen fluoride)
Hydrogen bonding (O–H···O)
Dispersion forces only
Hydrogen bonding (F–H···F)
Dipole-dipole forces (polar C–F bond, no H on F)
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Worked examples
Worked examples · reveal as you go
Worked example+5 XP on full reveal
Rank the following molecules in order of increasing boiling point and justify each position: (A) propane (C₃H₈, non-polar, MW 44), (B) propan-1-ol (C₃H₇OH, has O–H bond, MW 60), (C) chloromethane (CH₃Cl, polar, MW 51).
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Identify IMFs in each molecule: A (propane), no polar bonds, no N/O/F–H → only dispersion forces, MW = 44. B (propan-1-ol), has O–H bond → hydrogen bonding, MW = 60. C (chloromethane), polar C–Cl bond → dipole-dipole + dispersion, MW = 51.
Always check for H-bonding first (N–H, O–H, F–H). Then check for polarity. Then assess size for dispersion strength. This is the three-step framework.
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Rank IMF strength for each: A has dispersion only (weakest) → lowest BP. C has dispersion + dipole-dipole → moderate BP. B has dispersion + hydrogen bonding → highest BP.
A and C have similar size/electron counts, but C's extra dipole-dipole forces push it above A. B wins because H-bonding is far stronger than dipole-dipole, and B is also larger.
B's hydrogen bonding gives it a BP ~140°C higher than A, despite being only ~16 Da heavier. This shows how dramatically hydrogen bonding elevates BP compared to dispersion and dipole-dipole forces.
Worked example+5 XP on full reveal
The table shows BPs for Group 16 hydrides. H₂O (BP 100°C) breaks the trend seen in H₂S (−60°C), H₂Se (−41°C), H₂Te (−2°C). Explain the anomaly.
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Identify the general trend: H₂S → H₂Se → H₂Te shows BP increasing from −60°C to −41°C to −2°C. This is because MW increases down the group → more electrons → stronger dispersion forces → higher BP.
These molecules have only dispersion and weak dipole-dipole forces. The trend is smooth and predictable. If water followed the same trend (i.e. if it had no hydrogen bonding), extrapolation estimates its BP at roughly −75 to −90°C, the exact figure depends on how the trend is extrapolated, so treat it as a model-dependent estimate, not a measured value.
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Observe the H₂O anomaly: H₂O's actual BP is 100°C, roughly 180°C higher than the dispersion-only trend estimate (~−80°C).
This huge deviation needs explanation. The regular trend applies to S, Se, Te hydrides, but water is different. This always signals a special type of IMF.
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Identify the cause, Hydrogen bonding in H₂O: Oxygen is highly electronegative (χ = 3.5), making the O–H bond strongly polar. The δ+ hydrogen on one water molecule forms a hydrogen bond with the lone pair on the δ− oxygen of an adjacent molecule (O–H···O).
Hydrogen bonding is far stronger than the dispersion and dipole-dipole forces in H₂S, H₂Se, H₂Te. This explains the massive +180°C elevation.
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Explain why other hydrides don't H-bond: S, Se, and Te are not electronegative enough to create the large partial charge on H needed for hydrogen bonding. H₂S has only dispersion (from 34 electrons) + weak dipole-dipole → weaker total IMF → lower BP (−60°C).
Electronegativity is the deciding factor: χ(O) = 3.5 (H-bonds), χ(S) = 2.6 (no H-bonds). The 0.9-unit difference is enough to switch from hydrogen bonding to just dispersion/dipole-dipole.
Put the method in order
Sort the steps+7 XP
Click two steps to swap them. Put the IMF-prediction routine in the right order to rank the boiling points of CH₄, CH₃F and CH₃OH.
CH₃OH has an O–H bond, hydrogen bonding (the strongest IMF) is present.
For each molecule, check first whether H is bonded directly to N, O or F.
CH₄ has no N/O/F and is non-polar → only weak dispersion forces.
CH₃F has no H bonded to F (the H is on C) but the molecule is polar → dispersion + dipole-dipole.
Predicted BP order (lowest → highest): CH₄ < CH₃F < CH₃OH, matching observed −162°C < −78°C < 65°C.
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Misconception to fix
Common errors · the 3 traps that cost marks
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Misconception to fix
Wrong: Hardness and strength mean the same thing when describing materials.
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Misconception to fix
Right: Hardness refers to resistance to scratching or indentation. Strength refers to resistance to breaking under force. Diamond is extremely hard but brittle, it can be scratched by almost nothing yet shatters under impact. These are independent properties.
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Forgetting dispersion forces in polar molecules
Many students write that water "only has hydrogen bonds" or HCl "only has dipole–dipole". Dispersion forces act between every pair of molecules, they don't switch off just because a stronger IMF is also present. For large polar molecules, dispersion can even be the dominant contribution.
Fix: Always list dispersion forces alongside any dipole–dipole or hydrogen bonding when identifying IMFs in a molecule.
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Drill, then revisit
Quick-fire practice · 5 reps +2 XP per reveal
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List the three intermolecular forces from weakest to strongest.
Dispersion forces, dipole–dipole forces, and hydrogen bonding (N–H···N). All three IMFs operate.
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n-Pentane (BP 36 °C) and neopentane (BP 9.5 °C) have the same molecular formula C₅H₁₂. Why does n-pentane have the higher BP?
n-Pentane's straight-chain shape lets molecules contact each other along their full length, maximising dispersion forces. Neopentane is compact and branched, reducing contact area and weakening dispersion → lower BP.
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Why doesn't HCl form hydrogen bonds despite the H–Cl bond being polar?
Hydrogen bonding requires H bonded to F, O or N, atoms electronegative enough to put a large δ+ on H. Cl is not electronegative enough, so HCl only has dispersion + dipole–dipole forces.
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Predict the order of BP from lowest to highest: CH₄, CH₃F, CH₃OH. Justify each.
CH₄ < CH₃F < CH₃OH. CH₄ is non-polar, only weak dispersion (BP −162 °C). CH₃F is polar but the H is on C, so dispersion + dipole–dipole only (BP −78 °C). CH₃OH has an O–H bond → hydrogen bonding adds the strongest IMF (BP 65 °C).
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Revisit your thinking
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
The Atomic Structure tool shows that the atomic number of an element equals…
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Multiple choice
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next.
Fill the blanks+4 XP
Complete the explanation of water's anomalously high boiling point compared with H₂S. Type each missing word, then click Check.
Water contains O–H bonds. Oxygen is very , so each O–H bond is highly polar.
This lets H₂O molecules form strong bonds with each other.
H₂S cannot form these bonds because sulfur is not electronegative enough, only weak and weak dipole-dipole forces act between H₂S molecules.
Stronger IMFs in water → more energy needed → water boils at °C, while H₂S boils at −60°C.
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Short answer
ApplyApply4 MARKS
Q1. 6. Explain why the boiling point of water (100°C) is much higher than that of hydrogen sulfide (H₂S, −60°C), even though H₂S is a larger, heavier molecule. In your answer, identify all the IMFs present in each substance and explain which is stronger.
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Q2. 7. A student is given three substances with BP data: fluoromethane (CH₃F, BP −78°C), methanol (CH₃OH, BP 65°C), and methane (CH₄, BP −162°C). (a) Identify the IMFs present in each substance. (b) Explain the BP trend in terms of IMF strength.
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Q3. 8. Evaluate the statement: "The boiling point of a substance depends entirely on the type of intermolecular force present, a substance with hydrogen bonding will always have a higher boiling point than one with only dispersion forces." Is this statement correct? Provide evidence and reasoning to support your answer.
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📖 Comprehensive answers (click to reveal)
Activity 1
1. (a) CH₄: dispersion forces only (non-polar, no N/O/F–H bonds). (b) HCl: dispersion + dipole-dipole (polar C–Cl bond; no H-bonding as Cl not electronegative enough). (c) CH₃OH: dispersion + hydrogen bonding (O–H bond → O–H···O H-bonds). (d) NH₃: dispersion + hydrogen bonding (N–H bond → N–H···N H-bonds).
2. Ranking (lowest → highest): Butane (C₄H₁₀) < Dimethyl ether (CH₃OCH₃) < Ethanol (C₂H₅OH). Butane: non-polar → dispersion only; MW 58 but only dispersion → BP −1°C. Dimethyl ether: polar (C–O bonds) → dispersion + dipole-dipole; MW 46; BP −24°C. Despite being lighter, ether is higher than methane-sized molecules; but lower than ethanol. Ethanol: has O–H → hydrogen bonding is the dominant IMF; BP 78°C despite being same MW as ether (46), H-bonding is far stronger than dipole-dipole, explaining the ~100°C difference.
3. Student A is correct, but for the right reason: HF has the highest BP (19°C) among the hydrogen halides because F is the most electronegative atom, enabling strong F–H···F hydrogen bonding between HF molecules. HI has more electrons (larger dispersion forces) and its BP (−35°C) is higher than HBr (−67°C) and HCl (−85°C) but lower than HF. The hydrogen bonding in HF easily outweighs the increased dispersion forces in the larger, heavier HI. Student B's reasoning is partly correct (more electrons → stronger dispersion) but fails to account for the dominant H-bonding effect in HF.
Activity 2
A: BP increases steadily as chain length increases (from −162°C for CH₄ to 69°C for C₆H₁₄). The IMF responsible is dispersion forces, the only type present in non-polar alkanes. Each additional,CH₂, group adds approximately 8 electrons (6 from C + 2 from H bonds), increasing the size and polarisability of the electron cloud → stronger temporary dipoles → stronger dispersion forces → more energy to vaporise → higher BP.
B: Pentanol (BP 138°C) is 102°C higher than pentane (BP 36°C) despite only 16 Da difference in MW. Pentanol has an O–H group → it forms strong hydrogen bonds (O–H···O) between molecules, requiring much more energy to separate them. Pentane is non-polar with only dispersion forces, relatively weak. Hydrogen bonding in pentanol is approximately 5–10 times stronger than the dispersion forces in pentane, producing the ~100°C BP elevation.
C: Predicted BP ~98°C (actual = 98°C). Reasoning: each step in the series adds ~CH₂ (MW +14) and increases BP by approximately 25–35°C. From hexane (C₆H₁₄, BP 69°C), adding one CH₂ unit to give heptane (C₇H₁₆) should increase BP by ~28–30°C → predicted 97–99°C. Assumption: the trend is linear (constant increment per CH₂), which is approximately true for mid-range alkanes but slightly changes for very short or very long chains.
❓ Multiple Choice
NH₃ has N–H bonds → hydrogen bonding. HCl: H–Cl, no H-bonding. CH₄: no polar bonds, no H-bonding. CO₂: no N/O/F–H bonds (C=O bonds but no H attached to O).
Noble gases are monatomic and non-polar → only dispersion forces. Larger atoms have more electrons → stronger dispersion → higher BP.
CH₄: dispersion only, very small → lowest. C₂H₅Cl: dispersion + dipole-dipole → moderate. C₂H₅OH: dispersion + H-bonding → highest. Actual BPs: CH₄ −161°C, C₂H₅Cl 12°C, C₂H₅OH 78°C.
HF forms F–H···F hydrogen bonds due to F's very high electronegativity. HCl cannot H-bond (Cl insufficient electronegativity). H-bonding in HF dominates over the greater dispersion forces of larger HCl.
Same formula, same MW, same non-polar nature → same type of IMFs (dispersion only). The difference must be shape: n-pentane's linear form has more surface contact area → stronger total dispersion. Neopentane's compact sphere has minimal contact area → weaker dispersion. This shape effect on BP is an important concept.
Short Answer Model Answers
Q6 (4 marks): H₂O has dispersion forces AND hydrogen bonding (O–H···O). Oxygen's high electronegativity (χ = 3.5) makes O–H bonds strongly polar, creating a large δ+ on H, this H forms a hydrogen bond with the lone pair on O of an adjacent water molecule (1 mark). H₂S has only dispersion forces and weak dipole-dipole forces, sulfur (χ = 2.6) is not electronegative enough to enable hydrogen bonding, so no H···S hydrogen bonds form (1 mark). Hydrogen bonds in water are far stronger (5–10×) than the dispersion and dipole-dipole forces in H₂S, requiring much more energy to break on boiling (1 mark). Despite H₂S being larger and heavier (MW 34 vs 18), its weaker IMFs mean it boils at −60°C; water's strong H-bonding elevates its BP to 100°C, a difference of 160°C (1 mark).
Q7 (5 marks): CH₄ IMFs: dispersion forces only (non-polar, no polar bonds, no N/O/F–H) (1 mark). CH₃F IMFs: dispersion + dipole-dipole forces (C–F bond is polar: F is electronegative → permanent dipole; no H bonded to F, so no H-bonding) (1 mark). CH₃OH IMFs: dispersion + hydrogen bonding (O–H bond present; O is highly electronegative → H forms H-bonds with lone pairs on O of adjacent molecules) (1 mark). BP trend: CH₄ lowest (−162°C), weakest IMFs (dispersion only, small molecule) (1 mark). CH₃F is higher (−78°C), dipole-dipole forces in addition to dispersion, despite similar size to CH₄; polar C–F bond adds extra attractive force between molecules. CH₃OH highest (65°C), hydrogen bonding is far stronger than dipole-dipole, creating a much larger energy requirement to vaporise (1 mark).
Q8 (4 marks): The statement is incorrect as an absolute claim (1 mark). While hydrogen bonding is the strongest type of IMF, boiling point depends on the total strength of all IMFs, which is influenced by both the type of IMF AND the molecular size (number of electrons → dispersion force strength) (1 mark). A counterexample: hexadecane (C₁₆H₃₄, non-polar, BP 287°C) has only dispersion forces yet boils far higher than water (BP 100°C, which has hydrogen bonding). The very large dispersion forces from 130+ electrons in hexadecane dominate over the H-bonding in water's 10-electron molecule (1 mark). A more accurate statement would be: "For molecules of similar size, a substance with hydrogen bonding will generally have a higher boiling point than one with only dispersion forces. However, for molecules of very different sizes, the stronger dispersion forces in a much larger molecule can exceed the hydrogen bonding in a smaller molecule" (1 mark).
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Check what actually stuck
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quiz
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