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Carbon and its Compounds – NCERT Exercises

Q: How do I solve NCERT Class 10 Science Chapter 4 (Carbon and its Compounds) exercise questions for the CBSE board exam?

Solve NCERT Chapter 4 u2014 Carbon and its Compounds u2014 exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like carbon, hydrocarbons, functional groups, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step u2014 CBSE Class 10 board examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for the board exam.

Q: How many marks does Chapter 4 u2014 Carbon and its Compounds u2014 carry in the Class 10 Science CBSE paper?

Chapter 4 u2014 Carbon and its Compounds u2014 is part of the Class 10 Science syllabus and typically contributes 5u20139 marks in the CBSE board paper, depending on the annual weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like carbon, hydrocarbons, functional groups. Solving the NCERT exercises in this part is essential because CBSE directly references NCERT for question design.

🎓 Class 10 Science CBSE Theory Ch 4 — Carbon and its Compounds ⏱ ~22 min

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Chapter Summary — Carbon and its Compounds

1. Why carbon is special

Electronic configuration 2, 4 — four valence electrons.
Cannot easily lose or gain 4 e⁻; forms covalent bonds by sharing electrons.
Tetravalency (4 bonds per C) + catenation (C–C chains) → millions of compounds.
Allotropes: diamond (hard, 3D), graphite (layered, conductor), fullerene C₆₀, graphene, nanotubes.

2. Hydrocarbons & homologous series

Saturated — alkanes \(C_nH_{2n+2}\). Unsaturated — alkenes \(C_nH_{2n}\), alkynes \(C_nH_{2n-2}\).
Chains may be straight, branched, or ring (cyclic).
Homologous series — same general formula, members differ by –CH₂– (Δmass 14). Similar chemistry, graded physical properties.

3. Functional groups (quick reference)

Group	Formula	Suffix / prefix	Example
Halide	–X	chloro-, bromo- (prefix)	CH₃Cl
Alcohol	–OH	-ol	CH₃OH methanol
Aldehyde	–CHO	-al	CH₃CHO ethanal
Ketone	>C=O	-one	CH₃COCH₃ propanone
Carboxylic acid	–COOH	-oic acid	CH₃COOH ethanoic acid

4. Key reactions

Combustion: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{heat}\)

Oxidation: \(\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{alk.KMnO}_4} \text{CH}_3\text{COOH}\)

Addition (hydrogenation): \(\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3\text{CH}_3\)

Substitution: \(\text{CH}_4 + \text{Cl}_2 \xrightarrow{h\nu} \text{CH}_3\text{Cl} + \text{HCl}\)

Ester: \(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightleftharpoons[]{\text{H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\)

5. Ethanol, ethanoic acid & soaps — key facts

Ethanol (C₂H₅OH): colourless, miscible with water; reacts with Na (H₂↑); dehydrated to ethene by conc. H₂SO₄ at 443 K; harmful — causes liver damage, addiction.
Ethanoic acid (CH₃COOH): 5–8% in vinegar; pure form freezes at 17 °C (glacial acetic acid); reacts with base (salt + H₂O) and carbonates (CO₂↑); forms esters with alcohols.
Soap: Na salt of fatty acid. Has hydrophilic head + hydrophobic tail → forms micelles. Fails in hard water (forms scum with Ca²⁺/Mg²⁺).
Detergents: synthetic; work in hard water; some are non-biodegradable.

Keywords

Covalent bond

Lewis structure

Valence electrons

Tetravalency

Catenation

Allotropes

Diamond

Graphite

Fullerene (C₆₀)

Graphene

Hydrocarbon

Alkane

Alkene

Alkyne

Saturated

Unsaturated

Isomer

Homologous series

Functional group

IUPAC nomenclature

Combustion

Oxidation

Addition reaction

Substitution reaction

Hydrogenation

Esterification

Saponification

Ethanol

Ethanoic acid

Denatured alcohol

Glacial acetic acid

Ester

Soap

Detergent

Micelle

Scum

Hard water

Hydrophilic / hydrophobic

NCERT Textbook Exercises (pp. 71–72)

Q1 Ethane, with the molecular formula C₂H₆ has:
(a) 6 covalent bonds (b) 7 covalent bonds (c) 8 covalent bonds (d) 9 covalent bonds

Answer: (b) 7 covalent bonds. In H₃C–CH₃: one C–C bond + six C–H bonds = 7 covalent bonds.

Q2 Butanone is a four-carbon compound with the functional group:
(a) carboxylic acid (b) aldehyde (c) ketone (d) alcohol

Answer: (c) Ketone. The suffix "-one" denotes a ketone (>C=O). Butanone = CH₃COCH₂CH₃.

Q3 While cooking, if the bottom of the vessel is getting blackened on the outside, it means that:
(a) the food is not cooked completely (b) the fuel is not burning completely (c) the fuel is wet (d) the fuel is burning completely.

Answer: (b) The fuel is not burning completely. Incomplete combustion — due to insufficient air supply — deposits unburned carbon (soot) on the vessel, blackening it. Clean the burner and check the air holes.

Q4 Explain the nature of the covalent bond using the bond formation in CH₃Cl.

Solution: In CH₃Cl (chloromethane):

Carbon (2,4) has 4 valence electrons — needs 4 more to complete its octet.
Hydrogen has 1 valence electron — needs 1 more for its duplet.
Chlorine (2,8,7) has 7 valence electrons — needs 1 more for its octet.
Carbon shares one electron each with three hydrogens (three C–H single bonds) and one electron with a chlorine (one C–Cl single bond).
All bonds are single covalent bonds formed by mutual sharing of a pair of electrons. Each atom now has a stable noble-gas configuration.

Q5 Draw the electron dot structures for (a) ethanoic acid (b) H₂S (c) propanone (d) F₂.

Solution:
(a) Ethanoic acid CH₃COOH: H₃C–C(=O)–O–H. C of –COOH has a double bond (2 pairs) to one O and a single bond (1 pair) to –O–H. The =O has 2 lone pairs; the –OH oxygen has 2 lone pairs.
(b) H₂S: H:S:H — S shares one pair with each H (two single bonds), with 2 lone pairs on S.
(c) Propanone CH₃COCH₃: central C shares 2 pairs (double bond) with O and single bonds with each –CH₃. O has 2 lone pairs.
(d) F₂: F:F (single bond, 1 shared pair). Each F has 3 lone pairs.

Q6 What is a homologous series? Explain with an example.

Solution: A homologous series is a family of organic compounds with the same functional group and general formula, in which successive members differ by a –CH₂– unit (mass difference 14).
Example — alcohols: CH₃OH (methanol), C₂H₅OH (ethanol), C₃H₇OH (propanol), C₄H₉OH (butanol). General formula C_nH_2n+1OH. Chemical properties (e.g., reaction with Na) are similar; physical properties change gradually.

Q7 How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?

Solution:

Property	Ethanol (C₂H₅OH)	Ethanoic acid (CH₃COOH)
Smell	Mild, characteristic	Strong, pungent (vinegar-like)
Melting point	−114 °C	17 °C (freezes in winter — glacial)
Litmus test	Neutral — no effect	Turns blue litmus red (acidic)
With Na₂CO₃ / NaHCO₃	No reaction	Brisk effervescence (CO₂ released)
With NaOH	No reaction	Forms salt (sodium acetate) + water

The Na₂CO₃ test (effervescence) is the simplest chemical test.

Q8 Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Solution: Soap molecules have two ends — an ionic (hydrophilic) head and a long hydrocarbon (hydrophobic) tail. In water, the tails are repelled by water and cluster together, while the heads dissolve in water; this natural arrangement forms a spherical micelle, with tails inside and heads pointing outward.
In ethanol (an organic solvent), the hydrocarbon tails dissolve freely — there is no need for them to hide from the solvent. Hence no micelle forms in ethanol.

Q9 Why are carbon and its compounds used as fuels for most applications?

Solution: Carbon and its compounds are excellent fuels because:

They release a large amount of heat per unit mass on combustion (high calorific value).
They burn with little or no smoke in sufficient air — clean energy.
They have relatively low ignition temperatures (easy to light).
They are widely available (coal, petroleum, natural gas, wood).
Combustion products (CO₂, H₂O) are mostly non-toxic in open air.

Q10 Explain the formation of scum when hard water is treated with soap.

Solution: Hard water contains dissolved calcium (Ca²⁺) and magnesium (Mg²⁺) ions. When soap (sodium stearate) is added, these ions displace sodium and form insoluble calcium/magnesium stearate — a curdy white precipitate called scum.

\(2\text{C}_{17}\text{H}_{35}\text{COONa} + \text{Ca}^{2+} \rightarrow (\text{C}_{17}\text{H}_{35}\text{COO})_2\text{Ca}\downarrow + 2\text{Na}^+\)

Scum wastes soap and deposits on clothes, giving poor cleaning.

Q11 What change will you observe if you test soap with litmus paper (red and blue)?

Solution: Soap solution is slightly alkaline (sodium salt of a weak fatty acid). It will:

Turn red litmus blue.
Have no effect on blue litmus.

Q12 What is hydrogenation? What is its industrial application?

Solution: Hydrogenation is the addition of hydrogen (H₂) across the C=C (or C≡C) bonds of an unsaturated compound in the presence of a nickel (or platinum/palladium) catalyst, converting it to the corresponding saturated compound.

\(\text{R–CH=CH–R'} + \text{H}_2 \xrightarrow{\text{Ni}} \text{R–CH}_2\text{–CH}_2\text{–R'}\)

Industrial use: conversion of unsaturated vegetable oils (like sunflower, groundnut, cottonseed oils) into solid saturated fats — vanaspati ghee — by passing H₂ through the oil in the presence of a finely divided Ni catalyst.

Q13 Which of the following compounds will undergo addition reactions: C₂H₆, C₃H₈, C₃H₆, C₂H₂, CH₄?

Solution: Only unsaturated compounds undergo addition reactions. Checking each with \(C_nH_{2n+2}\):

C₂H₆ — saturated (alkane) → no
C₃H₈ — saturated → no
C₃H₆ — fits \(C_nH_{2n}\), unsaturated (propene) → yes
C₂H₂ — fits \(C_nH_{2n-2}\), unsaturated (ethyne) → yes
CH₄ — saturated → no

So C₃H₆ and C₂H₂ undergo addition reactions.

Q14 Give a test to distinguish between saturated and unsaturated hydrocarbons.

Solution: Bromine water test: Add a few drops of bromine water (reddish-brown) to each compound.

Unsaturated compounds (C=C or C≡C) quickly decolourise the bromine water by addition.
Saturated compounds (alkanes) do NOT react with bromine water in the dark; the colour remains.

Alternative test: saturated burns with clean blue flame; unsaturated burns with yellow sooty flame.

Q15 Explain the mechanism of the cleansing action of soaps.

Solution:

A soap molecule has a polar head (–COO⁻Na⁺, hydrophilic) and a long non-polar hydrocarbon tail (hydrophobic).
On a greasy cloth, the tails embed themselves into the oil/grease while the heads project into surrounding water.
Many soap molecules surround a single droplet of grease, forming a spherical micelle — tails inside, heads outside.
The outer shell of ionic heads makes the micelle water-soluble, so the grease droplet is effectively suspended in water.
On rinsing with water, the micelles (with trapped dirt) are carried away, and the cloth becomes clean.

Q16 Name the salts formed when (a) ethanoic acid reacts with NaOH, (b) dilute HCl reacts with NaOH, (c) ethanoic acid reacts with Na₂CO₃.

Solution:
(a) Sodium ethanoate (sodium acetate, CH₃COONa):

\(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\)

(b) Sodium chloride (NaCl):

\(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)

\(2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{CO}_2 + \text{H}_2\text{O}\)

Q17 Next homologue of CH₃OCH₃ (dimethyl ether) — give the name and formula.

Solution: Adding one –CH₂– unit to CH₃OCH₃ gives CH₃OC₂H₅ (methoxyethane / ethyl methyl ether, C₃H₈O). The molecular mass increases from 46 to 60.

Q18 Write the names of the next homologues of: (i) CH₃CH₂Br (ii) CH₃–CO–CH₃ (iii) CH₃CH₂CH₂OH.

Solution:

(i) CH₃CH₂Br (bromoethane) → next: CH₃CH₂CH₂Br (1-bromopropane)
(ii) CH₃COCH₃ (propanone) → next: CH₃COCH₂CH₃ (butanone)
(iii) CH₃CH₂CH₂OH (propan-1-ol) → next: CH₃CH₂CH₂CH₂OH (butan-1-ol)

Q19 Give the IUPAC names of: (a) CH₃CH(CH₃)CH₃ (b) CH₃COCH₂CH₂CH₂CH₃ (c) CH₃–CH=CH–CH₂–CH₃.

Solution:

(a) Longest straight chain has 3 carbons with a methyl branch on C2 → 2-methylpropane (isobutane).
(b) 6 carbons, ketone group on C2 → hexan-2-one.
(c) 5 carbons with a double bond between C2 and C3 → pent-2-ene.

Q20 How would you bring about the following conversions? Name the process and write the balanced chemical equation. (i) ethanol → ethanoic acid, (ii) ethanoic acid → ethyl ethanoate (ester).

Solution:
(i) Oxidation: Heat ethanol with alkaline KMnO₄ (or acidified K₂Cr₂O₇):

\(\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\Delta]{\text{alk. KMnO}_4} \text{CH}_3\text{COOH}\)

(ii) Esterification: Warm ethanoic acid with ethanol in the presence of a few drops of conc. H₂SO₄:

\(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightleftharpoons[]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\)

The ester has a characteristic sweet smell.

Q21 In the soap micelles, which end of the molecule is on the outer surface and which end is in the interior of the cluster?

Solution: In a soap micelle formed in water:

The ionic hydrophilic heads (–COO⁻Na⁺) are on the outer surface, interacting with water.
The long hydrophobic hydrocarbon tails are in the interior, trapping oil/grease.

Q22 What are oxidising agents? Name any two that can oxidise ethanol to ethanoic acid.

Solution: Oxidising agents are substances that donate oxygen (or remove hydrogen) to oxidise other substances. They are themselves reduced in the process.
Two oxidising agents used to oxidise ethanol to ethanoic acid:

Alkaline potassium permanganate (KMnO₄) — purple colour fades as it is reduced.
Acidified potassium dichromate (K₂Cr₂O₇) — orange colour turns green.

Q23 Would you be able to check if water is hard by using a detergent?

Solution: No. Detergents work effectively in both hard and soft water (they do not form scum with Ca²⁺/Mg²⁺) — lather forms equally well in both. Therefore a detergent cannot distinguish hard from soft water.
A soap should be used instead: in hard water, soap forms scum and gives little lather; in soft water, soap lathers easily.

Q24 You are given three test tubes containing: (i) ethanol, (ii) glucose, (iii) baking-soda (sodium hydrogen carbonate). Which test would you use to identify each?

Solution: Use a combination of simple tests:

Litmus test / Add ethanoic acid: Baking soda (NaHCO₃) reacts with CH₃COOH producing brisk effervescence (CO₂) — identifies baking soda.
Benedict's / Fehling's test: Glucose (a reducing sugar) gives a brick-red precipitate of Cu₂O on warming with Fehling's or Benedict's solution — identifies glucose.
Ethanol is the remaining liquid — confirm by adding a piece of Na metal: bubbles of H₂ gas are released (ester test / Na reaction).

Frequently Asked Questions — NCERT Exercises & Intext Questions

How do I solve NCERT Class 10 Science Chapter 4 (Carbon and its Compounds) exercise questions for the CBSE board exam?

Solve NCERT Chapter 4 — Carbon and its Compounds — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like carbon, hydrocarbons, functional groups, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 10 board examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for the board exam.

Are the NCERT intext questions from Carbon and its Compounds important for the Class 10 board exam?

Yes, NCERT intext questions for Chapter 4 Carbon and its Compounds are highly important for the CBSE Class 10 Science board exam. Many board questions are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — carbon, hydrocarbons, functional groups — that chapter-end questions build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE exam's primary source.

What types of questions from Carbon and its Compounds are asked in the CBSE Class 10 Science board exam?

The CBSE Class 10 board paper asks a mix of question types from Carbon and its Compounds: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of carbon, hydrocarbons, functional groups, ethanol. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.

How many marks does Chapter 4 — Carbon and its Compounds — carry in the Class 10 Science CBSE paper?

Chapter 4 — Carbon and its Compounds — is part of the Class 10 Science syllabus and typically contributes 5–9 marks in the CBSE board paper, depending on the annual weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like carbon, hydrocarbons, functional groups. Solving the NCERT exercises in this part is essential because CBSE directly references NCERT for question design.

Where can I find step-by-step NCERT solutions for Chapter 4 Carbon and its Compounds Class 10 Science?

You can find complete, step-by-step NCERT solutions for Chapter 4 Carbon and its Compounds Class 10 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about carbon, hydrocarbons, functional groups that examiners look for. This makes revision quick and exam-focused for Class 10 CBSE board students.

What is the best way to revise Carbon and its Compounds before the Class 10 Science board exam?

The best way to revise Carbon and its Compounds for the CBSE Class 10 Science board exam is a three-pass approach. First pass: skim the chapter and note down key terms like carbon, hydrocarbons, functional groups in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt previous CBSE board questions and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.

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