Homologous series: successive members and physical trends
Connect the shared group, general formula and CH₂ increment while keeping chemical similarity distinct from identical physical values.
Section 11.1 · Formulae, functional groups and terminology
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- State the Core homologous-series definition and required general formulas.
- Compare chemical properties with measured physical properties.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How does C₃H₈ differ from C₂H₆?
What you will learn
- Describe all five required homologous-series characteristics.
- Use supplied physical data and the CH₂ increment without inventing an exact next value.
The five characteristics work together
A homologous series has the same functional group and the same general formula. Successive members differ by a –CH₂– unit, display a trend in physical properties, and share similar chemical properties. The same general formula is a pattern with a variable, not a claim that every member has the same molecular formula.
In the alkane example, ethane C₂H₆, propane C₃H₈ and butane C₄H₁₀ add one C and two H at each step. With supplied C=12 and H=1, each CH₂ increment adds 14 to relative molecular mass. Similar characteristic reactions can coexist with different boiling points and molecular sizes.
A trend is not necessarily an equal numerical step
Within the stated simple-series comparison, boiling points generally rise with increasing chain length as intermolecular attractions become greater. This is a physical trend, while the shared characteristic reactions explain chemical similarity. Keep physical state changes separate from the formation of new substances in a reaction.
Supplied values can support a direction without establishing the exact next boiling point or a constant interval. Structural branching and the particular series also matter in wider comparisons; do not rank every organic compound solely by a count of carbon atoms. For this objective, state all five family characteristics, verify the CH₂ increment and read the actual given data.
Pause and explain
Which pairing can occur within a homologous series?
Worked example
Compare C₂H₆, C₃H₈ and C₄H₁₀ using C=12 and H=1. Explain the successive change and why identical boiling points are not required.
Show the worked solution
- Their relative molecular masses are 30, 44 and 58.
- Each successive formula adds CH₂ and each mass increases by 14.
- They can share similar chemical behaviour while physical properties follow a trend rather than staying identical.
Answer CH₂ increments and relative masses 30→44→58, with physical trends rather than identical values.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
State all five required homologous-series characteristics.
Give me a hint
Include structure, formula, successive difference, physical change and chemical similarity.
Compare my reasoning
- Members share a functional group and a general formula.
- Successive members differ by CH₂.
- They display a physical-property trend and similar chemical properties.
Same group; same general formula; CH₂ successive increment; physical-property trend; similar chemical properties.
Look for these in your work
- All five characteristics appear.
- Physical trends and chemical similarity are distinguished.
Using C=12 and H=1, compare the relative molecular masses of C₃H₈ and C₄H₁₀ and connect the difference to CH₂.
Give me a hint
Count both extra carbon and extra hydrogens.
Compare my reasoning
- C₃H₈ has relative molecular mass 3×12+8=44.
- C₄H₁₀ has 4×12+10=58.
- The difference 58−44=14 equals 12+2×1 for CH₂.
44 and 58; difference 14 from one CH₂ unit.
Look for these in your work
- Both masses use the given atom values.
- The increment matches the formula difference.
Hypothetical successive-member boiling points are 30°C, 60°C and 85°C. Describe the observed trend and assess a prediction of exactly 110°C for the next member.
Give me a hint
Compare both direction and intervals before making an exact forecast.
Compare my reasoning
- The supplied values increase with each member.
- The intervals are 30°C and 25°C, so an equal-step rule is not established.
- 110°C may be a proposed estimate, but the values do not determine it uniquely; further data/model information is needed.
Increasing trend; intervals 30°C then 25°C do not establish an exact next value of 110°C.
Look for these in your work
- The supported direction is identified.
- An unsupported exact extrapolation is not presented as certain.
Common mistakes
- Saying the same general formula means identical molecular formulas.
- Omitting physical trends or chemical similarity from the five characteristics.
- Replacing a CH₂ increment with CH₃.
- Inventing a constant boiling-point interval from a few supplied values.
What can you explain now?
What unit separates successive homologous-series members?
Compare with the explanation
A CH₂ unit.
The increase is one carbon and two hydrogens in the stated sequence.
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Make it stick
List the five characteristics, check a new CH₂ increment and describe a supplied physical trend without inventing its exact next measurement.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.