MP Board Class 10 Science Chapter 4: Carbon Compounds
MP Board Class 10 Science Chapter 4: Carbon and Its Compounds Important Questions —
Carbon and Its Compounds is one of the most important chapters in MP Board Class 10 Science, carrying 6–8 marks in the board exam every year. This chapter covers covalent bonding, versatile nature of carbon, homologous series, functional groups, soaps and detergents. Below are the most frequently asked questions with detailed answers, MCQs, PYQs, and FAQs to help you score full marks in the 2027 board exam.
📑 Table of Contents
- Covalent Bonding in Carbon
- Versatile Nature of Carbon — Catenation & Tetravalency
- Hydrocarbons — Saturated and Unsaturated
- Functional Groups & Homologous Series
- Ethanol, Ethanoic Acid & Cleansing Agents
- Multiple Choice Questions (1 Mark Each)
- Short Answer Questions (2–3 Marks)
- Long Answer Questions (5 Marks)
- Previous Year Questions (2018–2026)
- Frequently Asked Questions
⚛️ 1. Covalent Bonding in Carbon
Q1. What is a covalent bond? Explain with examples.
Answer: A covalent bond is a chemical bond formed between two atoms by mutual sharing of electrons. Carbon has 4 electrons in its outermost shell (electronic configuration 2,4). Instead of gaining or losing 4 electrons (which would require very high energy), carbon shares its 4 valence electrons with other atoms to achieve a stable octet.
Q2. Why does carbon form covalent bonds rather than ionic bonds?
Answer: Carbon has 4 valence electrons (electronic configuration 2,4). To form an ionic bond, carbon would need to either:
- Lose 4 electrons to become C⁴⁺ — this requires extremely high ionization energy (1st IE = 1086 kJ/mol, but 4th IE is very large), making it energetically unfavourable.
- Gain 4 electrons to become C⁴⁻ — this would require adding 4 electrons to a small nucleus, which is also difficult due to strong electron-electron repulsion.
Therefore, carbon attains stability by sharing its 4 valence electrons through covalent bonding — the most energy-efficient path.
Q3. Draw the electron dot structure of methane (CH₄), carbon dioxide (CO₂), and water (H₂O).
Answer:
🔗 2. Versatile Nature of Carbon — Catenation & Tetravalency
Q4. What is catenation? Why is carbon capable of catenation?
Answer: Catenation is the property of an element to form bonds with atoms of the same element to create long chains, branched chains, or rings. Carbon exhibits catenation to the maximum extent because:
- C—C bond is very strong (bond energy = 348 kJ/mol) — stable enough to form long chains.
- Small atomic size (atomic radius = 77 pm) allows the shared electrons to be held tightly between nuclei.
- Tetravalency — 4 valence electrons allow carbon to form bonds with up to 4 other carbon atoms, creating complex structures.
- Carbon can form single, double, and triple bonds with itself, leading to diverse structures.
Q5. Explain tetravalency of carbon and how it leads to formation of different types of compounds.
Answer: Tetravalency means carbon has 4 valence electrons and can form 4 covalent bonds with other atoms. This allows carbon to:
- Form chains — Straight chains (n-butane), branched chains (isobutane), and closed rings (cyclohexane).
- Form multiple bonds — C=C double bonds (alkenes) and C≡C triple bonds (alkynes).
- Bond with heteroatoms — N, O, S, Cl, etc., creating functional groups like —OH (alcohol), —COOH (carboxylic acid), —CHO (aldehyde).
- Create isomers — Same molecular formula, different structural arrangements (e.g., C₄H₁₀ has two isomers: n-butane and isobutane).
🧪 3. Hydrocarbons — Saturated and Unsaturated
Q6. Differentiate between saturated and unsaturated hydrocarbons with examples.
Q7. Explain isomerism with reference to butane (C₄H₁₀).
Answer: Isomerism is the phenomenon where compounds have the same molecular formula but different structural arrangements (different structures). Butane (C₄H₁₀) has two structural isomers:
- n-Butane (straight chain): CH₃—CH₂—CH₂—CH₃. Boiling point: −0.5°C.
- Isobutane (branched chain): CH₃—CH(CH₃)—CH₃ (or 2-methylpropane). Boiling point: −11.7°C.
Key observation: Branched isomers have lower boiling points due to reduced surface area, leading to weaker van der Waals forces.
🔬 4. Functional Groups & Homologous Series
Q8. What is a functional group? Give examples of important functional groups.
Answer: A functional group is an atom or group of atoms that determines the chemical properties of an organic compound. All compounds containing the same functional group show similar chemical behaviour regardless of the length of the carbon chain.
Q9. What is a homologous series? Write its characteristics.
Answer: A homologous series is a series of organic compounds having the same functional group, same general formula, and differing from each other by a —CH₂— unit (methylene group). For example, the alkane series: CH₄, C₂H₆, C₃H₈, C₄H₁₀…
Characteristics of homologous series:
- All members of a homologous series have the same functional group.
- They have the same general formula (e.g., CₙH₂ₙ₊₂ for alkanes).
- Each successive member differs by a —CH₂— group (14 u mass difference).
- Chemical properties are similar within the series (same functional group, same reactions).
- Physical properties show a gradual gradation — melting/boiling points increase as molecular mass increases.
Q10. What are addition and substitution reactions? Give examples.
Answer:
- Addition reaction: Unsaturated hydrocarbons add hydrogen, chlorine, bromine, or water across their double/triple bonds in the presence of catalysts (Ni, Pd, Pt).
Example: C₂H₄ + H₂ →(Ni, 200°C)→ C₂H₆ (Ethene to Ethane)
Example: C₂H₂ + 2Br₂ → C₂H₂Br₄ (Ethyne + Bromine → Tetrabromoethane, decolourization of bromine water) - Substitution reaction: A hydrogen atom in a saturated hydrocarbon is replaced by another atom (halogen) in the presence of sunlight.
Example: CH₄ + Cl₂ →(sunlight)→ CH₃Cl + HCl (Chloromethane formed)
🧴 5. Ethanol, Ethanoic Acid & Cleansing Agents
Q11. Write the properties and uses of ethanol.
Answer: Ethanol (C₂H₅OH) is a colourless liquid with a pleasant smell, commonly called alcohol.
Physical properties: Boiling point 78°C, miscible with water in all proportions (forms hydrogen bonds), neutral to litmus.
Chemical properties:
- Combustion: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + heat (clean blue flame).
- Oxidation to ethanoic acid: C₂H₅OH + 2[O] →(alkaline KMnO₄)→ CH₃COOH + H₂O.
- Dehydration to ethene: C₂H₅OH →(conc. H₂SO₄, 170°C)→ C₂H₄ + H₂O.
- Reaction with sodium: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑ (sodium ethoxide + hydrogen gas).
Uses: As a solvent in medicines (tinctures), in alcoholic beverages, as a fuel (mixed with petrol as gasohol), in manufacture of ethanoic acid and perfumes.
Q12. How is ethanoic acid (acetic acid) prepared from ethanol? Write its properties.
Answer: Ethanol is oxidized to ethanoic acid using alkaline potassium permanganate (KMnO₄) or potassium dichromate (K₂Cr₂O₇) as oxidizing agents.
Reaction: CH₃CH₂OH + 2[O] →(alk. KMnO₄)→ CH₃COOH + H₂O
Properties of ethanoic acid (CH₃COOH):
- Colourless liquid with pungent smell, boiling point 118°C.
- Freezes to a colourless crystalline solid (ice-like) at 17°C — called glacial acetic acid.
- Turns blue litmus red — acidic in nature.
- React with carbonates and bicarbonates to produce CO₂ gas:
2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂↑ - Reacts with ethanol in presence of conc. H₂SO₄ to form ester (esterification reaction):
CH₃COOH + C₂H₅OH ⇌(conc. H₂SO₄)⇌ CH₃COOC₂H₅ + H₂O (ethyl acetate — fruity smell)
Q13. Explain the mechanism of cleansing action of soaps and detergents.
Answer: Soaps are sodium or potassium salts of long-chain carboxylic acids (e.g., sodium stearate C₁₇H₃₅COONa). Each soap molecule has two parts:
- Hydrophilic (polar) head — COO⁻Na⁺ part, which is water-soluble (attracted to water).
- Hydrophobic (non-polar) tail — long hydrocarbon chain (C₁₇H₃₅—), which is oil-soluble (attracted to grease/dirt).
Cleansing mechanism:
- When soap is dissolved in water, the hydrophobic tails attach themselves to oil/dirt particles, while hydrophilic heads face outward toward water.
- This forms spherical structures called micelles — dirt trapped inside the hydrophobic core, surrounded by water-loving heads.
- Agitation (rubbing) helps break the dirt into smaller particles that get trapped in micelles.
- The micelles remain suspended in water (emulsification) and are washed away with water.
📝 6. Multiple Choice Questions (1 Mark Each)
MCQs from Carbon and Its Compounds appear frequently in the MP Board Class 10 Science paper. Practice these 10 MCQs to score full marks in the objective section.
- The general formula of alkanes is:
(a) CₙH₂ₙ (b) CₙH₂ₙ₊₂ ✓ (c) CₙH₂ₙ₋₂ (d) CₙH₂ₙ₊₁ - Which of the following shows catenation property to the maximum extent?
(a) Silicon (b) Sulphur (c) Carbon ✓ (d) Boron - The functional group in carboxylic acid is:
(a) —OH (b) —CHO (c) —COOH ✓ (d) —CO— - Ethene (C₂H₄) belongs to which homologous series?
(a) Alkanes (b) Alkenes ✓ (c) Alkynes (d) Alcohols - The reaction CH₄ + Cl₂ →(sunlight)→ CH₃Cl + HCl is an example of:
(a) Addition reaction (b) Substitution reaction ✓ (c) Combustion reaction (d) Esterification - Which reagent is used to convert ethanol to ethanoic acid?
(a) Conc. H₂SO₄ (b) Alkaline KMnO₄ ✓ (c) NaOH (d) HCl - What is the IUPAC name of CH₃COCH₃?
(a) Propanal (b) Propanone ✓ (c) Propanol (d) Propanoic acid - Which of the following is NOT a property of ethanoic acid?
(a) Turns blue litmus red (b) Freezes at 17°C (c) Reacts with Na₂CO₃ to give CO₂ (d) Has fruity smell ✓ - The number of covalent bonds in a methane molecule is:
(a) 2 (b) 3 (c) 4 ✓ (d) 5 - Soaps do not work well in hard water because they form:
(a) Micelles (b) Scum ✓ (c) Ester (d) Alcohol
✏️ 7. Short Answer Questions (2–3 Marks)
-
What are structural isomers? Draw the isomers of pentane (C₅H₁₂).
Structural isomers are compounds with the same molecular formula but different structural arrangements. Pentane has three isomers: n-pentane (CH₃CH₂CH₂CH₂CH₃), isopentane (CH₃CH(CH₃)CH₂CH₃), and neopentane (C(CH₃)₄). (2 marks) -
Explain esterification reaction with a chemical equation. Write one use of esters.
Esterification is the reaction between a carboxylic acid and an alcohol in the presence of conc. H₂SO₄ to form an ester (fruity-smelling compound). CH₃COOH + C₂H₅OH →(conc. H₂SO₄)→ CH₃COOC₂H₅ + H₂O. Esters are used in perfumes and flavouring agents. (2 marks) -
Why does carbon form compounds mainly by covalent bonding? Give two reasons.
(1) Carbon has 4 valence electrons — losing or gaining 4 electrons requires very high energy. (2) Small atomic size allows shared electron pairs to be held strongly between nuclei, forming stable covalent bonds. (2 marks) -
What is the difference between soaps and detergents? Write any two differences.
(1) Soaps are sodium salts of fatty acids; detergents are ammonium/sulphonate salts. (2) Soaps form scum in hard water; detergents do not. (3 marks) -
Write the electron dot structure of ethene (C₂H₄) and ethyne (C₂H₂).
Ethene: H₂C=CH₂ with each carbon sharing 2 electrons with the other carbon and 1 electron with each hydrogen. Ethyne: HC≡CH with each carbon sharing 3 electrons with the other. (2 marks) -
What happens when ethanol is heated with excess conc. H₂SO₄ at 170°C? Write the reaction.
Ethanol undergoes dehydration to form ethene (unsaturated hydrocarbon). C₂H₅OH →(conc. H₂SO₄, 170°C)→ C₂H₄ + H₂O. (2 marks) -
Give the IUPAC names of: (i) CH₃CH₂Br (ii) HCOOH (iii) CH₃CH₂CHO
(i) Bromoethane (ii) Methanoic acid (iii) Propanal. (3 marks) -
Why are carbon and its compounds used as fuels for most applications?
Carbon compounds (hydrocarbons) release a large amount of heat and light energy on combustion. They produce CO₂ and H₂O which are less harmful than other fuels. They are readily available and easy to transport as liquids (petrol, diesel, LPG). (2 marks)
📖 8. Long Answer Questions (5 Marks)
Q1. Describe the cleansing action of soaps with a neat diagram. Why do soaps not work well in hard water? How do detergents overcome this problem?
Answer: Soaps are sodium/potassium salts of long-chain fatty acids (e.g., sodium stearate C₁₇H₃₅COONa).
Cleansing action (step-by-step):
- Structure: Each soap molecule has a polar hydrophilic head (—COO⁻Na⁺) and a non-polar hydrophobic tail (long hydrocarbon chain).
- Micelle formation: In water, soap molecules cluster into spherical micelles — hydrophobic tails point inward (trapping oil/grease) and hydrophilic heads face outward (toward water).
- Emulsification: Dirt and grease get trapped in the hydrophobic core of micelles.
- Removal: Micelles remain suspended in water and are washed away, leaving the surface clean.
Problem with hard water: Hard water contains Ca²⁺ and Mg²⁺ ions. Soap reacts with these ions to form insoluble precipitates (scum/curd), wasting soap and leaving a residue on clothes.
2C₁₇H₃₅COONa + Ca²⁺ → (C₁₇H₃₅COO)₂Ca↓ + 2Na⁺
Detergent solution: Detergents are ammonium or sulphonate salts (e.g., sodium alkyl benzene sulphonate). Their calcium and magnesium salts are water-soluble, so they do not form scum in hard water and work effectively even in acidic conditions.
Q2. Explain the properties of ethanol and ethanoic acid. How can they be distinguished chemically?
Answer:
Q3. Explain homologous series with examples. Write the first five members of the alkane series and their properties.
Answer: A homologous series is a group of organic compounds with the same functional group and general formula, where each member differs from the next by a —CH₂— unit.
Alkane series (CₙH₂ₙ₊₂):
Key observations: As molecular mass increases, boiling point increases gradually. C₁–C₄ are gases, C₅–C₁₇ are liquids, and C₁₈+ are solids at room temperature. All alkanes show similar chemical properties (combustion, substitution reactions).
📋 Previous Year Questions (2018–2026)
The following questions from Carbon and Its Compounds have appeared in MP Board Class 10 Science exams over the past 8 years. Practice these to understand the exam pattern and marking scheme.
❓ Frequently Asked Questions
Q1. Is Carbon and Its Compounds a difficult chapter for MP Board Class 10?
No, with proper understanding of covalent bonding and regular practice of reaction equations, this chapter is easy to score. Focus on drawing structures and learning reactions.
Q2. How many marks does Carbon and Its Compounds carry in MP Board exam?
This chapter typically carries 6–8 marks in the MP Board Class 10 Science paper — 2-3 MCQs, one short answer (2-3 marks), and one long answer (5 marks).
Q3. Which topics are most important in this chapter for board exams?
Covalent bonding (electron dot structures), homologous series, functional groups, ethanol and ethanoic acid properties, esterification, and soap cleansing action are the most important topics.
Q4. What is the difference between ethanol and ethanoic acid?
Ethanol is a neutral alcohol (—OH group) with pleasant smell, while ethanoic acid is acidic (—COOH group) with pungent vinegar-like smell. Ethanol cannot turn blue litmus red, but ethanoic acid does.
Q5. What is the general formula of alkanes, alkenes, and alkynes?
Alkanes: CₙH₂ₙ₊₂, Alkenes: CₙH₂ₙ, Alkynes: CₙH₂ₙ₋₂. Alkanes have only single bonds, alkenes have one double bond, and alkynes have one triple bond.
Q6. How does carbon form millions of compounds?
Due to catenation (self-linking property) and tetravalency, carbon forms long chains, branched chains, rings, and bonds with heteroatoms, producing over 10 million known compounds.
Q7. What is saponification?
Saponification is the alkaline hydrolysis of esters to form soap and glycerol. This is the industrial process for soap manufacturing.
Q8. Why is bromine water test used in this chapter?
Bromine water test distinguishes saturated from unsaturated hydrocarbons. Unsaturated compounds decolourize bromine water (addition reaction), while saturated compounds do not.
Q9. What is glacial acetic acid?
Pure ethanoic acid (CH₃COOH) freezes to form ice-like crystals at 17°C — this solid form is called glacial acetic acid.
Q10. Are NCERT questions enough for this chapter in MP Board exams?
NCERT textbook questions are the foundation, but practising PYQs and additional questions from different formats (MCQs, assertion-reason, case-based) gives better exam readiness.
Q11. What is the best way to memorize carbon compound reactions?
Group reactions by functional group — all alcohols show similar reactions, all carboxylic acids show similar reactions. Write each reaction 3-5 times with proper balancing to build muscle memory.
Q12. Will questions on nano-materials or fullerenes come in the exam?
Only basic concepts from the NCERT textbook — diamond, graphite, buckminsterfullerene (C₆₀) — are part of the syllabus. Advanced nano-materials are not in the MP Board Class 10 Science syllabus.