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 C12.3 · Homologous series
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
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.
Sources for this lesson
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.
After trying it yourself, choose your next review. This is your self-assessment.
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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.