Chemistry • Year 11 • Module 1 • Lesson 10

Intermolecular Forces and Physical Properties

Build HSC Band 5–6 extended-response technique on evaluating IMF trends, designing investigations, and making evidence-based judgements about physical properties.

Master · Extended Response

1. Data + scenario: Group 16 hydride boiling points (Band 5–6)

8 marks Band 5–6

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)
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.”

Stuck? Plan: trend (H₂S→H₂Te: more electrons → stronger dispersion → higher BP) → extrapolate to predict H₂O BP ≈ −80 °C → actual 100 °C, anomaly ≈ +180 °C → O electronegativity (3.5) enables O–H···O H-bonds → evaluate statement (counterexample: hexadecane BP 287 °C > water 100 °C).

2. Experimental design — testing whether surface tension depends on IMF strength (Band 5–6)

7 marks Band 5–6

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: You have access to standard Year 11 laboratory equipment: a balance (±0.01 g), glass capillary tubes, a ruler, the three liquids (water, ethanol, hexane), and a dropper. The investigation must be completed 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 would allow a valid comparison of surface tension across the three liquids.
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.

Stuck? Consider: capillary rise (height of liquid up a thin glass tube) as a measurable proxy for surface tension. IV = liquid type; DV = capillary rise height (mm); controlled = capillary tube diameter, temperature. Predicted order: water > ethanol > hexane because H-bonds > H-bonds + dispersion > dispersion only.

Answers — Do not peek before attempting

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, which in turn will show greater capillary rise than hexane, because water (H-bonding) has stronger IMFs than ethanol (H-bonding + dispersion, but less extensive network) which has stronger IMFs than hexane (dispersion only). Independent variable: liquid type (water / ethanol / hexane). 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 three identical glass capillary tubes (same inner diameter, e.g. 0.5 mm) clamped vertically above three separate shallow dishes, each containing 20 mL of one liquid (water, ethanol, hexane respectively). (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 for each liquid using fresh liquid and note any evaporation (especially for hexane). Calculate the mean capillary rise height for each liquid and rank in order. [1 — four clear steps with a valid measurement strategy]

Falsification: If hexane or ethanol shows equal or greater capillary rise than water, 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. viscosity, density differences) are dominating the measurement [1].

Limitations: (1) Capillary rise depends on density and viscosity as well as surface tension; these differ between the three liquids, so rise height is not a pure measure of surface tension [1]. (2) Hexane is volatile and may partially evaporate during the experiment, reducing the liquid level and affecting the measurement [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/viscosity confound or volatility of hexane); 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).