CBSE Class 10 Science Carbon and its Compounds MCQs Set 11

Download CBSE MCQs for Class 10 Science: Chapter 04 Carbon and its Compounds

Review structured MCQ sets for Class 10 Science Chapter 04 Carbon and its Compounds. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chapter 04 Carbon and its Compounds

Access the complete set of multiple-choice questions for Chapter 04 Carbon and its Compounds below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Practice 35 more Class 10 Science MCQs from Carbon and Its Compounds, with answers, worked explanations and a teacher's note for every question. The questions cover why graphite conducts electricity, isomers of pentane and hexane, oxidation of ethanol and fermentation, and what is seen when soap is shaken with hard and soft water, and the page ends with 7 assertion-reason items.

Multiple Choice Questions

Question 1: Which of the following is not the use of graphite?
(a) It is used as lubricant
(b) It is used in manufacturing of lead-pencils
(c) It is used in manufacturing of artificial diamond
(d) It is used for making insulated plates
Show Answer & Explanation

Answer: (d) It is used for making insulated plates

Explanation:
1. Graphite conducts electricity, so it cannot be used to make insulating plates.
2. It is used as a lubricant, in pencil leads and as raw material for artificial diamond.

Teacher's Note:
1. Graphite is a conductor.
2. Check each use against its properties.

Question 2: Which two of these compounds are more likely to have similar properties? (P) CH3—CO—CH3 (Q) CH3—CH(OH)—CH3 (R) CH3—CH2—COOH (S) CH3—OH
(a) (P) and (Q)
(b) (Q) and (S)
(c) (P) and (R)
(d) (R) and (S)
Show Answer & Explanation

Answer: (b) (Q) and (S)

Explanation:
1. (Q) and (S) are both alcohols with the –OH group.
2. (P) is a ketone and (R) is a carboxylic acid, so they behave differently.

Teacher's Note:
1. Same functional group means similar properties.
2. Name the functional group in each compound.

Question 3: Graphite conducts electricity because
(a) it has free carbon atoms
(b) it has free electrons
(c) its one layer can slip over another layer
(d) None of these
Show Answer & Explanation

Answer: (b) it has free electrons

Explanation:
1. In graphite each carbon is bonded to three others, leaving one electron free.
2. These free electrons move and carry current.

Teacher's Note:
1. Diamond has no free electrons, so it does not conduct.
2. Slipping layers explain softness, not conduction.

Question 4: Number of free electron(s) in each carbon atom in graphite is/are
(a) two
(b) four
(c) one
(d) three
Show Answer & Explanation

Answer: (c) one

Explanation:
1. In graphite each carbon is bonded to three other carbon atoms, using three of its four valence electrons.
2. The fourth electron is free, which makes graphite a conductor.

Teacher's Note:
1. Free electrons explain electrical conduction.
2. In diamond all four electrons are used.

Question 5: How many single bonds are there in methane?
(a) Four
(b) Five
(c) Six
(d) Three
Show Answer & Explanation

Answer: (a) Four

Explanation:
1. Methane, CH4, has four C–H single bonds.
2. There is no other bond in the molecule.

Teacher's Note:
1. Carbon forms four covalent bonds.
2. Count the lines in the structure.

Question 6: The allotrope of carbon which is a good conductor of heat and electricity is:
(a) Diamond
(b) Graphite
(c) Charcoal
(d) None of these
Show Answer & Explanation

Answer: (b) Graphite

Explanation:
1. In graphite each carbon is joined to three others, leaving one free electron per atom.
2. These free electrons carry heat and electricity.

Teacher's Note:
1. Free electrons mean conduction.
2. Diamond conducts heat but not electricity.

Question 7: Consider the following statements regarding diamond and graphite: I. Diamond and graphite are isotopes of carbon. II. C–C bond length in diamond is greater than that in graphite. III. Graphite is poor conductor of electricity. Select the correct statement/s.
(a) I and II
(b) Only III
(c) Only II
(d) I, II and III
Show Answer & Explanation

Answer: (c) Only II

Explanation:
1. Diamond and graphite are allotropes (not isotopes), and graphite is a good conductor, so I and III are false.
2. The C–C bond is 154 pm in diamond and 141.5 pm in graphite, so II is true.

Teacher's Note:
1. Allotropes differ in structure, isotopes in neutrons.
2. Graphite conducts electricity.

Question 8: Observe the following pairs of organic compounds : (I) C4H9OH and C5H11OH (II) C7H15OH and C5H11OH (III) C6H13OH and C3H7OH. Which of these pair is a homologous series according to increasing order of carbon atom?
(a) (III) only
(b) (II) only
(c) (I) only
(d) All of these
Show Answer & Explanation

Answer: (c) (I) only

Explanation:
1. Only pair (I) has the second compound one –CH2– unit higher than the first, in increasing order of carbon atoms.
2. In pairs (II) and (III) the carbon number decreases, so they do not fit.

Teacher's Note:
1. Check the direction of increase in carbon atoms.
2. Adjacent members differ by CH2.

Question 9: How many structural isomers are possible for pentane (C5H12)?
(a) 1
(b) 4
(c) 2
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Pentane has three carbon skeletons: n-pentane, isopentane (2-methylbutane) and neopentane (2,2-dimethylpropane).
2. So there are 3 structural isomers.

Teacher's Note:
1. Draw the longest chain first, then shorten it and add branches.
2. Butane has 2 isomers, pentane has 3.

Question 10: In C6H14 the number of possible isomers is
(a) 3
(b) 6
(c) 4
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. The five isomers are n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.
2. So the answer is 5.

Teacher's Note:
1. Systematically shorten the main chain.
2. Do not count the same structure twice.

Question 11: The table shows six organic compounds A, B, C, D, E and F: A C7H16; B C8H16; C C4H6; D C6H10; E C5H10; F C9H20. Compound B is
(a) an alkane
(b) an alkene
(c) an alkyne
(d) none of these.
Show Answer & Explanation

Answer: (b) an alkene

Explanation:
1. C8H16 fits CnH2n with n = 8.
2. So compound B is an alkene.

Teacher's Note:
1. H = 2 × C means alkene.
2. Test n = 8.

Question 12: Among the following sets of compounds, choose the set having the same molecular formulae:
(a) Butane and iso-butane
(b) Cyclohexane and hexene
(c) Propanal and propanone
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Butane and iso-butane are C4H10, cyclohexane and hexene are C6H12, and propanal and propanone are C3H6O.
2. So all the pairs have the same molecular formula.

Teacher's Note:
1. Write the molecular formula of each compound.
2. Same formula, different structure means isomers.

Question 13: Structural formula of ethyne is
(a) H–C≡C–H
(b) H3–C≡C–H
(c) H2C=CH2
(d) H3C–CH3
Show Answer & Explanation

Answer: (a) H–C≡C–H

Explanation:
1. Ethyne (C2H2) has two carbon atoms joined by a triple bond, each also bonded to one hydrogen.
2. Option (b) has the wrong number of hydrogen atoms, (c) is ethene and (d) is ethane.

Teacher's Note:
1. Alkyne means a C≡C triple bond.
2. Each carbon must show four bonds.

Question 14: Which among the following is not an example of a saturated hydrocarbon?
(a) Cyclohexane
(b) Benzene
(c) Butane
(d) Propane
Show Answer & Explanation

Answer: (b) Benzene

Explanation:
1. Benzene has alternate C=C double bonds, so it is unsaturated.
2. Cyclohexane, butane and propane have only single bonds.

Teacher's Note:
1. Ring does not mean unsaturated.
2. Check for double bonds.

Question 15: The organic compound contains 4 carbon atoms, the root word according to IUPAC is:
(a) Meth–
(b) Eth–
(c) Prop–
(d) But–
Show Answer & Explanation

Answer: (d) But–

Explanation:
1. The root words are meth (1), eth (2), prop (3), but (4), pent (5) and hex (6).
2. Four carbon atoms give ‘but’.

Teacher's Note:
1. Memorise the root words up to ten carbons.
2. Butane has four carbons.

Question 16: When methane is burnt in an excess of air, the products of combustion are –
(a) C and H2O
(b) CO and H2O
(c) CO2 and H2
(d) CO2 and H2O
Show Answer & Explanation

Answer: (d) CO2 and H2O

Explanation:
1. With excess air methane burns completely: CH4 + 2O2 → CO2 + 2H2O + heat.
2. Carbon dioxide and water are the products.

Teacher's Note:
1. Complete combustion gives CO2 and H2O.
2. Limited air gives CO or soot.

Question 17: A student conducted an activity, where he took a naphthalene ball and burned it. He observed that it gives a yellow flame with lots of black smoke and sooty deposits around it. What type of hydrocarbon does naphthalene contain?
(a) Unsaturated, as black smoke represents complete combustion.
(b) Unsaturated, as sooty deposit represents unburnt hydrocarbons.
(c) Saturated, as it gives a yellow flame which represents complete combustion.
(d) Saturated, as the burning of any substance represents a complete combustion.
Show Answer & Explanation

Answer: (b) Unsaturated, as sooty deposit represents unburnt hydrocarbons.

Explanation:
1. A yellow sooty flame shows incomplete combustion and unburnt carbon.
2. Unsaturated compounds burn this way because of their high carbon content.

Teacher's Note:
1. Saturated compounds burn with a clean blue flame in enough air.
2. Soot means unburnt carbon.

Question 18: CH3–CH2–OH → CH3–COOH, in the presence of alkaline KMnO4 and heat. In the above given reaction, alkaline KMnO4 acts as
(a) reducing agent
(b) oxidising agent
(c) catalyst
(d) dehydrating agent
Show Answer & Explanation

Answer: (b) oxidising agent

Explanation:
1. Ethanol gains oxygen and is converted into ethanoic acid, so it is oxidised.
2. The substance that oxidises another is the oxidising agent, here alkaline KMnO4.

Teacher's Note:
1. Addition of oxygen means oxidation.
2. Acidified K2Cr2O7 can do the same job.

Question 19: When ethanoic acid is heated with NaHCO3 the gas evolved is –
(a) H2
(b) CO2
(c) CH4
(d) CO
Show Answer & Explanation

Answer: (b) CO2

Explanation:
1. CH3COOH + NaHCO3 → CH3COONa + H2O + CO2.
2. Carbon dioxide is evolved with brisk effervescence.

Teacher's Note:
1. Carboxylic acids release CO2 from carbonates.
2. The gas turns lime water milky.

Question 20: A & B both compounds give H2 gas with sodium. If A & B react in presence of acid catalyst then they form ethyl acetate. Thus, A & B would be -
(a) CH4, CH3OH
(b) HCOOH, CH3COOH
(c) CH3COOH, C2H5OH
(d) C3H7COOH, C3H7OH
Show Answer & Explanation

Answer: (c) CH3COOH, C2H5OH

Explanation:
1. Both an acid and an alcohol give H2 with sodium.
2. Ethyl acetate is formed from ethanoic acid and ethanol in the presence of an acid catalyst.

Teacher's Note:
1. Ester = acid + alcohol.
2. Name the ester to find its parent acid and alcohol.

Question 21: An organic liquid ‘A’ with acidified potassium dichromate gave product ‘B’. The compound ‘B’ on heating with methanol in presence of concentrated sulphuric acid formed compound ‘C’ which on subsequent treatment with sodium hydroxide formed two product ‘D’ and ‘E’. The product ‘D’ is known to affect the optic nerve causing blindness. Intake of ‘D’ in very small quantities can cause death. What are compound ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’?
(a) A = Ethanol, B = Ethanoic acid, C = Methanol, D = Sodium acetate, E = Methyl ethanoate
(b) A = Ethanol, B = Ethanoic acid, C = Methyl ethanoate, D = Methanol, E = Sodium acetate
(c) A = Sodium acetate, B = Ethanoic acid, C = Methyl ethanoate, D = Methanol, E = Ethanol
(d) A = Ethanol, B = Ethanoic acid, C = Methyl ethanoate, D = Sodium acetate, E = Methanol
Show Answer & Explanation

Answer: (b) A = Ethanol, B = Ethanoic acid, C = Methyl ethanoate, D = Methanol, E = Sodium acetate

Explanation:
1. Ethanol is oxidised to ethanoic acid, which reacts with methanol to give the ester methyl ethanoate.
2. On treatment with NaOH the ester gives methanol (D, the poisonous alcohol) and sodium ethanoate (E).

Teacher's Note:
1. Poisonous alcohol means methanol.
2. Follow the chain step by step.

Question 22: Fermentation of sugarcane juice produces
(a) Ethanol
(b) Ethanal
(c) Acetic acid
(d) Gluconic acid
Show Answer & Explanation

Answer: (a) Ethanol

Explanation:
1. Yeast enzymes convert the sugar in sugarcane juice into ethanol and carbon dioxide.
2. So ethanol is the product.

Teacher's Note:
1. Yeast is used in fermentation.
2. CO2 is released as a by-product.

Question 23: A compound Y on treatment with alkaline KMnO4 followed by acidification gives a compound X. Compound X also reacts with compound Y in presence of few drops of H2SO4 to form a sweet smelling compound Z. The compound X, Y and Z are, respectively.
(a) Ethanoic acid, Sodium ethanoate, Ethane
(b) Ethanoic Acid, Ethanol, Ethylethanoate
(c) Ethanoic Acid, Ethanal, Ethene
(d) Ethanol, Ethanoic Acid, Sodium Ethanoate
Show Answer & Explanation

Answer: (b) Ethanoic Acid, Ethanol, Ethylethanoate

Explanation:
1. Oxidation of ethanol (Y) by alkaline KMnO4 gives ethanoic acid (X).
2. X and Y react in the presence of H2SO4 to form the sweet-smelling ester ethyl ethanoate (Z).

Teacher's Note:
1. Alcohol oxidises to carboxylic acid.
2. Acid + alcohol gives ester.

Question 24: Vinegar is a solution of
(a) 50%-60% acetic acid in alcohol
(b) 5%-8% acetic acid in alcohol
(c) 5%-8% acetic acid in water
(d) 50%-60% acetic acid in water
Show Answer & Explanation

Answer: (c) 5%-8% acetic acid in water

Explanation:
1. Vinegar is a dilute aqueous solution containing about 5–8% ethanoic acid.
2. The solvent is water, not alcohol.

Teacher's Note:
1. Pure ethanoic acid is called glacial acetic acid.
2. Vinegar is used as a food preservative.

Question 25: The molecular formula of an organic compound ‘X’ is C2H6O. It is an active ingredient of all alcoholic drinks. A small piece of Na is dropped into the test tube containing ‘X’ as a result new compound ‘Y’ is formed with the evolution of colourless and odourless gas. Identify the compound 'Y'.
(a) CH3COOH
(b) CH3CH2OH
(c) CH3CH2ONa
(d) CH3CH2CH2ONa
Show Answer & Explanation

Answer: (c) CH3CH2ONa

Explanation:
1. X is ethanol, which reacts with sodium: 2C2H5OH + 2Na → 2C2H5ONa + H2.
2. So Y is sodium ethoxide, CH3CH2ONa, and the gas is hydrogen.

Teacher's Note:
1. Alcohol + Na gives sodium alkoxide + H2.
2. The gas burns with a pop sound.

Question 26: The observations recorded by the first student on adding a few drops of soap solution to 10 mL distilled water and hard water in separate test tubes is given. Select the correct observation:
(a) Distilled water: Lot of foam observed; Hard water: Lot of foam observed
(b) Distilled water: Curdy precipitate observed; Hard water: Lot of foam observed
(c) Distilled water: Lot of foam observed; Hard water: Curdy precipitate observed
(d) Distilled water: Curdy precipitate observed; Hard water: Curdy precipitate observed
Show Answer & Explanation

Answer: (c) Distilled water: Lot of foam observed; Hard water: Curdy precipitate observed

Explanation:
1. Distilled water has no calcium or magnesium salts, so soap gives plenty of foam.
2. Hard water has Ca2+ and Mg2+ ions, which form a curdy precipitate (scum) with soap.

Teacher's Note:
1. Hard water contains calcium and magnesium salts.
2. Scum is the curdy precipitate.

Question 27: A student studies that soap solution results in micelle formation which helps to remove dirt. It has a unique orientation which helps in keeping the dirt out of the water. What helps the dirt to rinse away?
(a) Suspension of the dirt in the micelles.
(b) A collection of water molecules in the centre of the micelle.
(c) Attraction between the ionic end and the dirt to remove it.
(d) Mixing of the soap molecules along with the dirt to make it heavier.
Show Answer & Explanation

Answer: (a) Suspension of the dirt in the micelles.

Explanation:
1. The hydrocarbon tails of soap molecules dissolve in the oil and dirt at the centre of the micelle.
2. The ionic heads face the water, so the micelle with the trapped dirt is rinsed away.

Teacher's Note:
1. Tails are oil-loving, heads are water-loving.
2. Dirt sits inside the micelle.

Question 28: Observe the table: A – 10 mL hard water + soap (5 drops): white curd like scum formed; B – 15 mL hard water + detergent (5 drops): white curd like scum formed; C – 8 mL hard water + soap (5 drops): lot of lather formed; D – 12 mL hard water + detergent (5 drops): lot of lather formed. Which among the given test tubes give correct result?
(a) A & B
(b) C & D
(c) A & D
(d) B & D
Show Answer & Explanation

Answer: (c) A & D

Explanation:
1. In hard water soap forms scum with little lather (A is correct, C is wrong).
2. Detergent forms plenty of lather in hard water (D is correct, B is wrong).

Teacher's Note:
1. Soap fails in hard water; detergent works.
2. Test each row separately.

Assertion–Reason Questions

Question 29:
Assertion (A): Diamond and graphite are allotropes of carbon.
Reason (R): Some elements can have several different structural forms in the same physical state. These differing forms are called allotropes.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Diamond and graphite are two solid structural forms of carbon.
2. R gives the definition of allotropes, so it explains A.

Teacher's Note:
1. Same element, same state, different structure.
2. Fullerene is a third allotrope.

Question 30:
Assertion (A): Graphite is slippery to touch.
Reason (R): The various layers of carbon atoms in graphite are held together by weak van der Waals forces.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Layers in graphite are held by weak van der Waals forces and slide over each other.
2. So R explains why graphite is slippery.

Teacher's Note:
1. Weak forces between layers allow sliding.
2. This property makes graphite a dry lubricant.

Question 31:
Assertion (A): Following are the members of a homologous series: CH3OH, CH3CH2OH, CH3CH2CH2OH
Reason (R): A series of compounds with same functional group but differing by –CH2– unit is called a homologous series.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. All three have the –OH group and differ by one –CH2– unit.
2. This is exactly the definition of a homologous series, so R explains A.

Teacher's Note:
1. Look for the same functional group and a CH2 difference.
2. Write the next member to check your understanding.

Question 32:
Assertion (A): Olefins have the general formula CnH2n+1.
Reason (R): There is at least one double bond between two carbon atoms in their molecules.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Olefins (alkenes) have the general formula CnH2n, so A is false.
2. R is true, since alkenes contain a C=C double bond.

Teacher's Note:
1. Olefin is another name of alkene.
2. CnH2n+1 is an alkyl group.

Question 33:
Assertion (A): CH3Cl is obtained from CH4 by the action of Cl2 in the presence of sunlight.
Reason (R): It is obtained by addition reaction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Methane reacts with chlorine in sunlight to give CH3Cl, so A is true.
2. The reaction is a substitution, not an addition, because methane is saturated, so R is false.

Teacher's Note:
1. Saturated compounds undergo substitution.
2. Unsaturated compounds undergo addition.

Question 34:
Assertion (A): Ethanol is first member of the alcohol homologous series.
Reason (R): A homologous series can be represented by a general formula.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. The first member of the alcohol series is methanol, so A is false.
2. R is true, as every series has a general formula.

Teacher's Note:
1. Methanol is first, ethanol is second.
2. Know the first member of each series.

Question 35:
Assertion (A): Soaps are not effective in hard water.
Reason (R): The charged ends of detergents do not form insoluble precipitates with the calcium and magnesium ions in hard water.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Soap reacts with Ca2+ and Mg2+ ions of hard water to form insoluble scum, so A is true.
2. R is also true, but it describes detergents and does not explain why soaps fail.

Teacher's Note:
1. Soap forms scum, detergent does not.
2. R is about detergents, not soaps.

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Practice MCQs for Class 10 Science Chapter 04 Carbon and its Compounds

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