Chemistry • Year 11 • Module 1 • Lesson 10
Intermolecular Forces and Physical Properties
Build HSC advanced extended-response technique on evaluating IMF trends, designing investigations, and making evidence-based judgements about physical properties.
1. Data + scenario: Group 16 hydride boiling points (Evaluate)
8 marks Evaluate
Scenario. A student is analysing the boiling points of the Group 16 hydrides to understand the role of intermolecular forces. The table below shows the data. The student expects the boiling point to increase smoothly down the group as molecular mass increases, but notices something unexpected about water.
| Hydride | Molecular mass (g mol−1) | Boiling point (°C) | IMFs expected from trend |
|---|---|---|---|
| H2O | 18 | 100 | |
| H2S | 34 | −60 | |
| H2Se | 81 | −41 | |
| H2Te | 130 | −2 |
Data from NIST WebBook (illustrative values consistent with published literature).
Q1. Analyse and evaluate the boiling point data for the Group 16 hydrides and assess the role of intermolecular forces in explaining both the general trend and the anomaly. In your response you must:
- Complete the “IMFs expected from trend” column in the table above for H2S, H2Se, and H2Te.
- Describe the general trend in boiling point from H2S to H2Te and explain it quantitatively in terms of IMF type and molecular size.
- Identify the anomaly and use Pauling electronegativity values to explain precisely why H2O does not follow the trend.
- Estimate what the boiling point of H2O would be if it had only dispersion and dipole-dipole forces (by extrapolating the H2S–H2Te trend), and quantify the anomaly in °C.
- Evaluate the statement: “The type of IMF always matters more than molecular size in determining boiling point.”
2. Experimental design, testing whether surface tension depends on IMF strength (Evaluate)
7 marks Evaluate
Research question. Surface tension is a physical property that arises from intermolecular forces; stronger IMFs should produce higher surface tension. A student hypothesises that water will have a higher surface tension than ethanol and hexane, because water forms hydrogen bonds while the others have weaker IMFs. Design an investigation to test this hypothesis.
Constraints (safety): You may use water and ethanol hands-on (ethanol with ventilation and no ignition sources). Hexane is not used in this open experiment (too flammable and volatile), instead take its published surface-tension value from a data table as a secondary data point. Equipment: a balance (±0.01 g), glass capillary tubes, a ruler, a dropper, and a surface-tension reference table. Complete in one laboratory session (50 min).
Q2. Design the investigation and present it in the format below.
- State your hypothesis as a testable prediction (include the independent and dependent variables).
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe a procedure in at least four numbered steps that allows a valid comparison of surface tension (water and ethanol measured directly; hexane taken from published data), and explain how you account for capillary rise also depending on density, tube radius and contact angle, not surface tension alone.
- State what result would falsify the hypothesis and what it would mean for the IMF model.
- Identify two limitations of the design and suggest one improvement to increase reliability.
Q1, Sample Band 6 response (8 marks), annotated
IMF column completion: H2S: dispersion + dipole-dipole (S not electronegative enough for H-bonding; χ(S) = 2.6). H2Se: dispersion + dipole-dipole (Se: χ ≈ 2.4). H2Te: dispersion + dipole-dipole (Te: χ ≈ 2.1, even weaker dipole).
General trend (H2S to H2Te): Boiling point increases from −60 °C (H2S) to −41 °C (H2Se) to −2 °C (H2Te) [1]. This is because molecular mass increases down the group (34 → 81 → 130 g mol−1) → more electrons → larger, more polarisable electron clouds → stronger dispersion forces → more energy required to separate molecules on boiling [1].
H2O anomaly and electronegativity explanation: H2O is the anomalous molecule. It is the smallest (MW 18) yet has the highest boiling point (100 °C) [1]. Oxygen has an exceptionally high Pauling electronegativity (χ = 3.5), making the O–H bond strongly polar and placing a large δ+ charge on hydrogen. This allows each H2O molecule to form up to two O–H···O hydrogen bonds with adjacent molecules. Sulfur (χ = 2.6), selenium, and tellurium are far less electronegative and cannot create the necessary δ+ on H for hydrogen bonding to occur; their boiling points are governed only by dispersion and weak dipole-dipole forces [1].
Extrapolation and quantifying the anomaly: Extrapolating the linear H2S–H2Se–H2Te trend backward to MW 18 predicts H2O should boil at approximately −75 to −80 °C [1]. The actual boiling point is 100 °C, giving an anomaly of approximately +175 to +180 °C. This +180 °C elevation is entirely attributable to hydrogen bonding [1].
Evaluation of the statement: The statement is an oversimplification [1]. While IMF type is crucial for molecules of similar size (as seen in H2O vs H2S), molecular size can override IMF type for very large molecules. A counterexample: hexadecane (C16H34, non-polar, dispersion forces only, BP 287 °C) has a higher boiling point than water (BP 100 °C, hydrogen bonding), because its 130+ electrons generate dispersion forces stronger than water’s H-bonds. A more accurate statement is that “both IMF type and molecular size determine boiling point; for similar sizes, IMF type dominates, but for vastly different sizes, dispersion forces can override H-bonding.” [1]
Marking criteria (8 marks): 1 = correctly identifies IMFs for H2S, H2Se, H2Te (no H-bonding); 1 = describes trend with correct direction and links to molecular mass / electron count / dispersion forces; 1 = correctly identifies H2O as anomalous with a BP of 100 °C; 1 = explains anomaly using Pauling electronegativity of O (3.5) and the requirement for O–H bond + lone pair on adjacent O; 1 = extrapolates trend to obtain estimated BP ≈ −75 to −85 °C; 1 = correctly quantifies anomaly as ≈180 °C and attributes it to H-bonding; 1 = evaluates the statement as an oversimplification with a valid counterexample (e.g. hexadecane); 1 = uses precise terminology throughout (Pauling electronegativity, dispersion forces, permanent dipole, polarisability, H-bond donor/acceptor).
Q2, Sample Band 6 response (7 marks), annotated
Hypothesis: If surface tension is proportional to IMF strength, then water will show greater capillary rise than ethanol, and both will exceed hexane (whose surface tension is taken from published data), because water (H-bonding) has stronger IMFs than ethanol (H-bonding + dispersion) which has stronger IMFs than hexane (dispersion only). Independent variable: liquid type (water and ethanol measured directly; hexane from data). Dependent variable: height of capillary rise (mm). Controlled variables: capillary tube inner diameter (same tube each time), liquid temperature (room temperature ≈ 20 °C), volume of liquid in container. [1 , hypothesis with IV and DV]
Procedure: (1) Set up two identical glass capillary tubes (same inner diameter, e.g. 0.5 mm) clamped vertically above two shallow dishes containing 20 mL of water and 20 mL of ethanol respectively (ethanol used with ventilation and no ignition sources). (2) Lower each capillary tube so its base just touches the liquid surface and allow the liquid to rise for 2 minutes. (3) Measure and record the height (mm) of liquid rise in each capillary using a ruler, reading from the bottom of the meniscus. (4) Repeat steps 1–3 three times per liquid using fresh liquid; calculate the mean rise for water and ethanol, then place hexane in the ranking using its published surface-tension value. Because capillary rise also depends on density, tube radius and contact angle, keep the tube and temperature constant and treat rise as a proxy, not an absolute surface-tension measurement. [1 , four clear steps with a valid measurement strategy]
Falsification: If ethanol shows equal or greater capillary rise than water, or if hexane’s published surface tension does not rank lowest, the hypothesis is falsified [1]. This would imply that IMF strength does not correlate with surface tension, which would challenge the IMF model for physical properties or suggest confounding factors (e.g. density, tube radius, contact angle) are dominating the measurement [1].
Limitations: (1) Capillary rise depends on density, tube radius and contact angle as well as surface tension, so rise height is not a pure measure of surface tension [1]. (2) Using a published value for hexane (rather than measuring it) assumes the reference conditions match the experiment; differences in temperature or purity would affect comparability [1].
Improvement: Use a more precise surface tension measurement method, such as counting drops from a standardised dropper (drop weight method): more drops per mL indicates stronger surface tension. Repeat all measurements at least five times per liquid to improve reliability [1].
Marking criteria (7 marks): 1 = testable hypothesis naming IV (liquid type) and DV (capillary rise / surface tension measure); 1 = four clear procedure steps including a valid quantitative measurement; 1 = states what would falsify the hypothesis and what it would mean for the IMF model; 1 = identifies one valid limitation (density/tube-radius/contact-angle confound, or the assumption in using a published hexane value); 1 = identifies a second valid limitation; 1 = proposes a specific, realistic improvement; 1 = uses precise terminology (H-bonding, dispersion forces, independent/dependent/controlled variable, surface tension, IMF).