Class 10 Science Chapter 4 Carbon and its Compounds Practice Paper 4

Class: X Subject: Science Session: 2025-26 Chapter: 04 - Carbon and its Compounds Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

SECTION A - Multiple Choice Questions (1 mark each)

Q1. The electron dot structure of nitrogen molecule shows:
(a) A single bond between two nitrogen atoms
(b) A double bond between two nitrogen atoms
(c) A triple bond between two nitrogen atoms
(d) No bond between nitrogen atoms

Q2. Which of the following hydrocarbons undergo addition reactions?
(a) C₂H₆
(b) C₃H₈
(c) C₃H₆
(d) CH₄

Q3. Ethanol on oxidation with alkaline KMnO₄ gives:
(a) Methanol
(b) Methanal
(c) Ethanoic acid
(d) Ethane

Q4. The correct structural formula of butanone is:
(a) CH₃-CH₂-CH₂-CHO
(b) CH₃-CO-CH₂-CH₃
(c) CH₃-CH₂-COOH
(d) CH₃-CH₂-CH₂-CH₃

Q5. Which of the following is used to convert ethanol into ethene?
(a) Alkaline KMnO₄
(b) Conc. H₂SO₄ at 443 K
(c) Dilute HCl
(d) Water

Q6. Carbon forms four covalent bonds because it has:
(a) 2 electrons in outermost shell
(b) 4 electrons in outermost shell
(c) 6 electrons in outermost shell
(d) 8 electrons in outermost shell

Q7. The smell of esters is:
(a) Pungent like vinegar
(b) Sweet and fruity
(c) Rotten eggs
(d) Odorless

Q8. The difference in molecular mass between successive members of a homologous series is:
(a) 12 u
(b) 14 u
(c) 16 u
(d) 18 u

Q9. Scum formation is due to:
(a) Reaction of soap with soft water
(b) Reaction of soap with hard water containing Ca²⁺ and Mg²⁺ ions
(c) Reaction of detergent with hard water
(d) Micelle formation

Q10. In graphite, each carbon atom is bonded to how many other carbon atoms?
(a) One
(b) Two
(c) Three
(d) Four

SECTION B - Short Answer Questions (2 marks each)

Q11. What is the role of concentrated sulphuric acid in esterification reaction? Write the equation for the formation of ethyl ethanoate.

Q12. Write two tests to distinguish between ethanol and ethanoic acid.

Q13. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?

Q14. Draw the structures of the following compounds: (i) Propanone (ii) Propanal

SECTION C - Short Answer Questions (3 marks each)

Q15.
(a) What are hydrocarbons? Give examples.
(b) Classify hydrocarbons based on the nature of carbon-carbon bonds.
(c) Write the general formula for each class.

Q16. Three students A, B and C prepared soap solutions by dissolving equal amounts of soap in hard water, soft water and distilled water respectively. (i) In which case will the soap work most effectively? (ii) In which case will scum formation be maximum? (iii) Explain the reason for your answers.

Q17. Write balanced chemical equations for the following reactions: (i) Ethene reacts with hydrogen in presence of nickel catalyst (ii) Methane burns in presence of oxygen (iii) Ethanoic acid reacts with sodium hydrogencarbonate

SECTION D - Long Answer Question (5 marks)

Q18.
(a) Explain why carbon exhibits the property of catenation to a much greater extent than silicon, even though both are in the same group of the periodic table.
(b) What are the two features seen in carbon that give rise to a large number of carbon compounds?
(c) Write the molecular formula and draw the structures of: (i) Ethyne (ii) Benzene (iii) Cyclopentane
(d) What are alkanes, alkenes and alkynes? Write one example of each.

SECTION E - Case Study Based Questions (4 marks each)

Q19. Case Study 1: Covalent Bonding in Carbon Compounds Carbon has four electrons in its outermost shell and needs four more electrons to achieve noble gas configuration. It could gain four electrons forming C⁴⁻ anion, but it would be difficult for the nucleus with six protons to hold ten electrons. It could lose four electrons forming C⁴⁺ cation, but it would require a large amount of energy. Therefore, carbon overcomes this problem by sharing its valence electrons with other atoms through covalent bonding. The shared electrons belong to the outermost shells of both atoms. Carbon forms single, double or triple bonds with other carbon atoms or with atoms of other elements.

Answer the following questions: (i) Why is it difficult for carbon to form C⁴⁺ ion? (1 mark) (ii) What is a covalent bond? (1 mark) (iii) How many bonds are present in a molecule of ethane (C₂H₆)? (1 mark) (iv) Name the type of bond present in N₂ molecule. (1 mark)

Q20. Case Study 2: Functional Groups and Nomenclature In a hydrocarbon chain, one or more hydrogens can be replaced by elements such as oxygen, nitrogen, sulphur or halogens. The element replacing hydrogen is called a heteroatom. These heteroatoms confer specific properties to the compound and are called functional groups. The presence of a functional group decides the properties of the carbon compound regardless of the length of the carbon chain. For example, all alcohols have similar chemical properties because they contain the -OH functional group.

Answer the following questions: (i) What is a functional group? (1 mark) (ii) Name the functional group present in CH₃COOH. (1 mark) (iii) Write the general formula of alcohols. (1 mark) (iv) What is the IUPAC name of CH₃CH₂CH₂CHO? (1 mark) DETAILED ANSWER KEY - PAPER 04

SECTION A - Answers to MCQs

Q1.
(c) A triple bond between two nitrogen atoms Explanation: Nitrogen (atomic number 7) has 5 valence electrons. To achieve noble gas configuration, each nitrogen shares 3 electrons with another nitrogen atom, forming a triple bond (N≡N).

Q2.
(c) C₃H₆ Explanation: C₃H₆ (Propene) is an unsaturated hydrocarbon containing a double bond. Only unsaturated hydrocarbons undergo addition reactions. C₂H₆, C₃H₈, and CH₄ are saturated hydrocarbons.

Q3.
(c) Ethanoic acid Explanation: C₂H₅OH + [O] → CH₃COOH (using alkaline KMnO₄ or acidified K₂Cr₂O₇). Ethanol is oxidized to ethanoic acid.

Q4.
(b) CH₃-CO-CH₂-CH₃ Explanation: Butanone has 4 carbon atoms with a ketone group (C=O) on the second carbon. Structure: CH₃-CO-CH₂-CH₃.

Q5.
(b) Conc. H₂SO₄ at 443 K Explanation: C₂H₅OH → C₂H₄ + H₂O (using conc. H₂SO₄ at 443 K). This is a dehydration reaction where water is removed from ethanol.

Q6.
(b) 4 electrons in outermost shell Explanation: Carbon (atomic number 6) has electronic configuration 2, 4. It has 4 valence electrons and forms 4 covalent bonds to achieve stable octet configuration.

Q7.
(b) Sweet and fruity Explanation: Esters are sweet-smelling substances with fruity odors. They are used in making perfumes and as flavoring agents.

Q8.
(b) 14 u Explanation: Successive members of a homologous series differ by -CH₂- unit. Molecular mass of - CH₂- = 12 + 2(1) = 14 u.

Q9.
(b) Reaction of soap with hard water containing Ca²⁺ and Mg²⁺ ions Explanation: Soap reacts with calcium and magnesium ions in hard water to form insoluble calcium and magnesium salts (scum), reducing cleaning efficiency.

Q10.
(c) Three Explanation: In graphite, each carbon atom is bonded to three other carbon atoms in the same plane, forming hexagonal layers. The fourth electron is free to move.

SECTION B - Answers to Short Answer Questions

Q11. Role of concentrated sulphuric acid in esterification reaction and equation. Answer: Role of Concentrated Sulphuric Acid: • Acts as a catalyst to speed up the esterification reaction • Acts as a dehydrating agent by removing water formed during the reaction • This helps shift the equilibrium forward, increasing ester formation Equation for formation of Ethyl Ethanoate: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (Ethanoic acid) + (Ethanol) → (Ethyl ethanoate/Ester) + (Water) [Conc. H₂SO₄, Heat]

Q12. Write two tests to distinguish between ethanol and ethanoic acid. Answer: Test 1: Litmus Test • Ethanol: Does not affect litmus paper (neither red nor blue litmus changes) • Ethanoic acid: Turns blue litmus paper red (acidic nature) Test 2: Sodium Carbonate/Sodium Hydrogencarbonate Test • Ethanol: No reaction occurs, no gas evolved • Ethanoic acid: Brisk effervescence occurs, CO₂ gas is evolved Chemical equation: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂↑ or CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑ The CO₂ gas can be tested by passing it through lime water, which turns milky.

Q13. Why is the conversion of ethanol to ethanoic acid an oxidation reaction? Answer: The conversion of ethanol to ethanoic acid is an oxidation reaction because: Chemical Equation: C₂H₅OH + [O] → CH₃COOH (Ethanol) + (Oxygen from oxidizing agent) → (Ethanoic acid) [Using alkaline KMnO₄ or acidified K₂Cr₂O₇] Reasons:

1. Addition of oxygen: • Oxygen is added to ethanol molecule • -CH₂OH group is converted to -COOH group • This involves addition of oxygen atom

2. Removal of hydrogen: • Hydrogen atoms are removed from ethanol • C₂H₅OH (6 H atoms) → CH₃COOH (4 H atoms)

3. Definition of oxidation: • Oxidation is defined as addition of oxygen or removal of hydrogen • Both processes occur in this reaction • Hence, it is an oxidation reaction

Q14. Draw the structures of (i) Propanone (ii) Propanal Answer: (i) Propanone (CH₃COCH₃): H O H | || | H — C — C C — H | || | H H or simply CH₃ — CO — CH₃ (Ketone group on middle carbon) (ii) Propanal (CH₃CH₂CHO): H H O | | || H — C — C — C | | \ H H H or simply CH₃ — CH₂ — CHO (Aldehyde group at terminal carbon)

SECTION C - Answers to Short Answer Questions

Q15.
(a) What are hydrocarbons?
(b) Classify hydrocarbons
(c) General formula
(a) Hydrocarbons: Hydrocarbons are organic compounds containing only carbon and hydrogen atoms. Examples: • Methane (CH₄) • Ethene (C₂H₄) • Ethyne (C₂H₂) • Benzene (C₆H₆)
(b) Classification of Hydrocarbons based on carbon-carbon bonds:

1. Saturated Hydrocarbons (Alkanes): • Contain only single bonds between carbon atoms • All carbon atoms are bonded to maximum number of hydrogen atoms • Less reactive • Example: Methane (CH₄), Ethane (C₂H₆)

2. Unsaturated Hydrocarbons: (i) Alkenes: • Contain one or more carbon-carbon double bonds (C=C) • More reactive than alkanes • Example: Ethene (C₂H₄), Propene (C₃H₆) (ii) Alkynes: • Contain one or more carbon-carbon triple bonds (C≡C) • Most reactive among hydrocarbons • Example: Ethyne (C₂H₂), Propyne (C₃H₄)
(c) General Formula for each class: Class General Formula Example (n=2) Alkanes CₙH₂ₙ₊₂ C₂H₆ (Ethane) Alkenes CₙH₂ₙ C₂H₄ (Ethene) Alkynes CₙH₂ₙ₋₂ C₂H₂ (Ethyne)

Q16. Soap solutions in hard water, soft water and distilled water - effectiveness and scum formation Answer: (i) Most effective soap solution: Student C (who prepared soap solution in distilled water) will find soap working most effectively. Student B (soft water) will also get good results, but distilled water is purest. (ii) Maximum scum formation: Student A (who used hard water) will observe maximum scum formation. (iii) Explanation: Hard Water (Student A): • Contains dissolved calcium (Ca²⁺) and magnesium (Mg²⁺) ions • Soap reacts with these ions to form insoluble calcium and magnesium salts • 2C₁₇H₃₅COONa + Ca²⁺ → (C₁₇H₃₅COO)₂Ca + 2Na⁺ (Soap) (Insoluble scum) • Scum is a white curdy precipitate • A lot of soap is wasted in scum formation • Soap cannot form lather until all Ca²⁺ and Mg²⁺ are removed • Least effective cleaning Soft Water (Student B):

• Contains very few or no Ca²⁺ and Mg²⁺ ions • Little or no scum formation • Soap works effectively • Forms lather easily • Good cleaning action Distilled Water (Student C): • Completely free from all dissolved salts and impurities • No scum formation at all • Soap works most effectively • Forms lather immediately • Best cleaning action

Q17. Write balanced chemical equations for given reactions Answer: (i) Ethene reacts with hydrogen in presence of nickel catalyst: C₂H₄ + H₂ → C₂H₆ (Ethene) + (Hydrogen) → (Ethane) [Ni catalyst, Heat] • This is an addition reaction (hydrogenation) • Unsaturated ethene is converted to saturated ethane • Used in conversion of vegetable oils to vegetable ghee (ii) Methane burns in presence of oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light (Methane) + (Oxygen) → (Carbon dioxide) + (Water) • This is a combustion reaction (oxidation) • Complete combustion in sufficient oxygen • Produces CO₂, H₂O, heat and light • Clean blue flame is observed (iii) Ethanoic acid reacts with sodium hydrogencarbonate:

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑ (Ethanoic acid) + (Sodium hydrogencarbonate) → (Sodium ethanoate) + (Water) + (Carbon dioxide) • Brisk effervescence occurs • CO₂ gas is evolved • Salt (sodium ethanoate) and water are formed • This reaction is used to test for carboxylic acids

SECTION D - Answer to Long Answer Question

Q18.
(a) Why carbon shows more catenation than silicon
(b) Two features of carbon
(c) Structures
(d) Alkanes, alkenes, alkynes
(a) Why carbon exhibits catenation to a greater extent than silicon: Although both carbon and silicon belong to Group 14 and have 4 valence electrons, carbon shows much greater catenation due to:

1. Atomic Size: • Carbon has smaller atomic radius (77 pm) compared to silicon (118 pm) • Smaller size allows stronger orbital overlap • Nucleus can hold shared electrons more strongly

2. Bond Strength: • C-C bond energy = 347 kJ/mol (very strong) • Si-Si bond energy = 226 kJ/mol (weaker) • Stronger C-C bonds are more stable • Silicon compounds are more reactive and less stable

3. Chain Length: • Carbon can form very long stable chains (thousands of atoms) • Silicon forms chains of only 7-8 atoms maximum • Silicon-hydrogen compounds (silanes) are highly reactive Therefore, carbon shows maximum catenation among all elements.
(b) Two features of carbon that give rise to large number of compounds:

1. Catenation: • Unique ability to form bonds with other carbon atoms • Forms long chains, branched chains, or rings • Carbon atoms can be linked by single, double or triple bonds • C-C bond is very strong and stable

2. Tetravalency: • Carbon has 4 valence electrons • Can form 4 covalent bonds • Can bond with 4 other carbon atoms or other elements • Forms compounds with H, O, N, S, Cl and many other elements • Each carbon can bond in multiple ways These two properties together enable carbon to form millions of compounds.
(c) Molecular formula and structures: (i) Ethyne (C₂H₂): H — C ≡ C — H (Triple bond between two carbon atoms) (ii) Benzene (C₆H₆): H | H—C C—H ║ ║ H—C C—H | H (Hexagonal ring with alternating double bonds) (iii) Cyclopentane (C₅H₁₀):

H₂C — CH₂ / \ H₂C CH₂ \ / C H₂ (Pentagon/five-membered ring)
(d) Alkanes, Alkenes and Alkynes: Type Definition General Example Formula Propane Saturated hydrocarbons with only single bonds between (C₃H₈) Alkanes CₙH₂ₙ₊₂ carbon atoms CH₃-CH₂- CH₃ Propene Unsaturated hydrocarbons with one or more double (C₃H₆) Alkenes CₙH₂ₙ bonds between carbon atoms CH₃- CH=CH₂ Propyne Unsaturated hydrocarbons with one or more triple (C₃H₄) Alkynes CₙH₂ₙ₋₂ bonds between carbon atoms CH₃- C≡CH

SECTION E - Answers to Case Study Based Questions

Q19. Case Study 1: Covalent Bonding in Carbon Compounds (i) Why is it difficult for carbon to form C⁴⁺ ion? It is difficult for carbon to form C⁴⁺ ion because: • Carbon would need to lose 4 electrons from its outermost shell • This requires a very large amount of energy (high ionization energy) • After losing 4 electrons, only 2 electrons remain with 6 protons in nucleus • The nucleus cannot stably hold such a small electron cloud • Therefore, formation of C⁴⁺ is energetically very unfavorable (ii) What is a covalent bond?

A covalent bond is a chemical bond formed by the sharing of electron pairs between two atoms. Features: • Electrons are shared, not transferred • Shared electrons belong to both atoms • Both atoms achieve stable electronic configuration • Strong bonds within molecule, weak forces between molecules • Forms when atoms have similar electronegativity Example: H-H bond in H₂ molecule (each H shares one electron) (iii) How many bonds are present in ethane (C₂H₆)? Total number of bonds in ethane = 7 bonds Structure: H₃C-CH₃ • 1 C-C single bond (between two carbon atoms) • 6 C-H single bonds (3 from each carbon) • Total = 1 + 6 = 7 covalent bonds (iv) Type of bond present in N₂ molecule:

Triple covalent bond (N≡N) • Each nitrogen has 5 valence electrons • Each nitrogen needs 3 more electrons to complete octet • Both nitrogen atoms share 3 pairs of electrons • Forms very strong triple bond • This makes N₂ very stable and unreactive

Q20. Case Study 2: Functional Groups and Nomenclature (i) What is a functional group? A functional group is an atom or group of atoms that replaces hydrogen in a hydrocarbon and determines the characteristic chemical properties of the compound. Key points: • Confers specific properties to the compound • Properties remain same regardless of carbon chain length • Same functional group gives similar chemical behavior • Different from just substituents - they define chemical class Examples: -OH (alcohol), -CHO (aldehyde), -COOH (carboxylic acid) (ii) Functional group present in CH₃COOH:

Carboxylic acid group (-COOH) or Carboxyl group Structure of functional group: O || — C — O — H It consists of: • Carbonyl group (C=O) • Hydroxyl group (-OH) • Together forming -COOH group (iii) General formula of alcohols: CₙH₂ₙ₊₁OH or CₙH₂ₙ₊₂O where n = number of carbon atoms Examples: • n=1: CH₃OH (Methanol) • n=2: C₂H₅OH (Ethanol) • n=3: C₃H₇OH (Propanol) All contain the -OH (hydroxyl) functional group attached to a carbon chain. (iv) IUPAC name of CH₃CH₂CH₂CHO: Butanal Explanation: • But- indicates 4 carbon atoms • -al is the suffix for aldehyde functional group (-CHO) • Aldehyde group is always at terminal position • Structure: CH₃-CH₂-CH₂-CHO (Also called Butyraldehyde in common nomenclature)