Science — CBSE Class 10 board question

(i) Homologous Series:
A homologous series is a series of carbon compounds having the same functional group, similar chemical properties, and the same general formula, in which successive members differ by a –CH₂– unit (14 mass units).

(ii) As the number of carbon atoms increases in a homologous series, the molecular mass increases. Greater molecular mass means stronger intermolecular forces of attraction, requiring more energy to overcome them. Hence, C₄H₈ (4 carbons) has higher melting and boiling points than C₃H₆ or C₂H₄.

(iii) All members of a homologous series have the same functional group, which determines chemical properties. Since the functional group remains unchanged throughout the series, there is no gradation in chemical properties.

(iv)

• Aldehyde (Propanal): CH₃–CH₂–CHO

Structure: $\text{CH}_3\text{-CH}_2\text{-}{\overset{O}{\overset{\|}{C}}}\text{-H}$

• Ketone (Propanone): CH₃–CO–CH₃

Structure: $\text{CH}_3\text{-}{\overset{O}{\overset{\|}{C}}}\text{-CH}_3$

Source: Chapter 4 – Carbon and its Compounds, Sections 4.2, 4.2.3

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• (i) The definition must include: same functional group, same general formula, differ by –CH₂–, similar chemical properties. All four earn full marks.
• (ii) The key logic is: more carbons → higher molecular mass → stronger intermolecular forces → higher boiling/melting point.
• (iii) Examiners want the reason: the functional group (not the chain) governs chemical properties, and it stays the same throughout the series.
• (iv) Both IUPAC names (Propanal, Propanone) and correct structural formulae are expected. The aldehyde –CHO is at the end of the chain; the ketone carbonyl (C=O) is between two carbon atoms. Draw clearly or write the condensed structural formula.