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(i) Define a homologous… (Class 10 Science 2024, Q34A)

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ScienceCBSE Class 10 · Board exam 2024, set 31/1/1 · 5 marks · Long answer

CBSE Class 10 Science 2024: question 34(A)

Set in the CBSE Class 10 board exam of 2024, paper 31/1/1, section D, question 34(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(A) · 5 marks · Long answerCarbon and its Compounds

(i) Define a homologous series of carbon compounds.
(ii) Why is the melting and boiling points of higher than that of or ?
(iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ?
(iv) Write the name and structures of (i) aldehyde and (ii) ketone with molecular form .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Think about what stays the same and what changes from one member of a series to the next.

Hint 2

For C₃H₆O, place the C=O at the end of the chain for the aldehyde and in the middle for the ketone.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) A homologous series is a series of carbon compounds that have the same functional group and similar chemical properties. Successive members differ by a unit (14 u in mass).

    1 mark

  2. (ii) has a higher molecular mass than and . In a homologous series, melting and boiling points increase as molecular mass increases.

    1 mark

  3. (iii) Chemical properties depend on the functional group. All members of a homologous series have the same functional group, so their chemical properties remain similar and show no gradation.

    1 mark

  4. (iv)(i) Aldehyde: Propanal, .

    ½ mark

  5. Structure: , i.e. the C=O is at the end of the chain bonded to an H.

    ½ mark

  6. (iv)(ii) Ketone: Propanone (acetone), .

    ½ mark

  7. Structure: , i.e. the C=O is on the middle carbon.

    ½ mark

Answer: (i) Same functional group, similar properties, differ by CH₂; (ii) higher molecular mass; (iii) same functional group decides properties; (iv) propanal CH₃CH₂CHO, propanone CH₃COCH₃

Where the marks are

  • Definition of homologous series (1 mark)
  • Melting/boiling point rises with molecular mass (1 mark)
  • Same functional group, so similar chemical properties (1 mark)
  • Propanal name and structure (½ + ½ marks)
  • Propanone name and structure (½ + ½ marks)

Revise the chapter

Practise this chapterBoard-style questions from Carbon and its Compounds, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 31/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.