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NCERT Class 10 Science — Chapter 4

Carbon and its Compounds

Covalent bonding (tetravalency, catenation), allotropes of carbon (diamond, graphite, fullerenes), saturated vs unsaturated hydrocarbons, homologous series, functional groups, and chemical reactions of Ethanol and Ethanoic acid (esterification, saponification).

Quick Key Takeaways:
Versatile Nature of Carbon: Carbon forms millions of compounds due to: (1) Tetravalency (valency 4, shares 4 electrons), (2) Catenation (unique ability to form long carbon-carbon covalent chains and rings).
Hydrocarbon Classifications: Alkanes (CnH2n+2C_n H_{2n+2} - single bonds, saturated), Alkenes (CnH2nC_n H_{2n} - double bond, unsaturated), Alkynes (CnH2n−2C_n H_{2n-2} - triple bond, unsaturated).
Homologous Series: A family of organic compounds having the same functional group and similar chemical properties, where consecutive members differ by a −CH2−-\text{CH}_2- unit and 1414 u molecular mass.
Reactions of Ethanol & Ethanoic Acid: Esterification (Ethanol+Ethanoic Acid→conc. H2SO4Ester+H2O\text{Ethanol} + \text{Ethanoic Acid} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{Ester} + \text{H}_2\text{O}); Saponification (Ester+NaOH→Soap+Alcohol\text{Ester} + \text{NaOH} \rightarrow \text{Soap} + \text{Alcohol}); Cleansing action of soaps (micelles with hydrophilic ionic head and hydrophobic hydrocarbon tail).
Chemistry Equation & Reaction HelperDeconstruct combustion, ethanol oxidation, esterification, and saponification
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Jump to Concept:1. 1. Covalent Bonding, Allotropes & Homologous Series Architecture2. 2. Chemical Reactions: Ethanol, Ethanoic Acid & Micelle Cleansing3. 3. High-Yield Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. Practical Laboratory & Competency-Based Case Drill: Esterification & Scented Ester Synthesis (NCERT Activity 4.8)5. 5. CBSE Examiner Marking Scheme, Scientific Notation & Deduction Traps
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1. Covalent Bonding, Allotropes & Homologous Series Architecture

Fundamental Principles

Core scientific laws, chemical equations, anatomical structures, and visual model for Carbon and its Compounds.

• Why Carbon Forms Covalent Bonds (Neither C4+C^{4+} nor C4−C^{4-})
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Carbon has atomic number 66 (electronic configuration: 2,42, 4).
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Why it cannot gain 44 electrons (C4−C^{4-}): The nucleus with 6 protons would be unable to hold onto 10 electrons (severe electrostatic repulsion).
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Why it cannot lose 44 electrons (C4+C^{4+}): It would require an enormous amount of ionization energy to remove 4 electrons from a tiny atom.
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Solution: Carbon shares its valence electrons with other carbon or other element atoms to form covalent bonds.
• Allotropes of Carbon
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Diamond: Rigid 3D tetrahedral network of carbon atoms. Hardest known natural substance; non-conductor of electricity (no free electrons).
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Graphite: Hexagonal planar layers held by weak van der Waals forces. Smooth, slippery lubricant; good conductor of electricity due to free delocalized π\pi-electrons.
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Fullerenes (Buckminsterfullerene C60C_{60}): Carbon atoms arranged in the shape of a football (geodesic dome with 20 hexagons and 12 pentagons).
📊 Carbon Compounds: Covalent Bonding & Functional Group ArchitectureVisual Model
1. Alkane (Single)CₙH₂ₙ₊₂ · Methane (CH₄)CHHHHSaturated (Single bond)2. Alkene (Double)CₙH₂ₙ · Ethene (C₂H₄)CCHHHHUnsaturated (Double)3. Alkyne (Triple)CₙH₂ₙ₋₂ · Ethyne (C₂H₂)HCCHUnsaturated (Triple)

Visual schematic mapping carbon tetravalency, catenation chains, homologous series, and the esterification-saponification cycle.

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2. Chemical Reactions: Ethanol, Ethanoic Acid & Micelle Cleansing

Mechanisms & Experiments

Step-by-step chemical reaction mechanisms, experimental activities, and physiological pathways for Carbon and its Compounds.

• Reactions of Ethanol & Ethanoic Acid
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1. Combustion: CH4+2O2⟶CO2+2H2O+Heat+Light\text{CH}_4 + 2\text{O}_2 \longrightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{Heat} + \text{Light}
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2. Oxidation of Ethanol to Ethanoic Acid:
CH3CH2OH→Alkaline KMnO4+Δ or Acidified K2Cr2O7+ΔCH3COOH\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Alkaline } \text{KMnO}_4 + \Delta \text{ or Acidified } \text{K}_2\text{Cr}_2\text{O}_7 + \Delta} \text{CH}_3\text{COOH}
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3. Addition Reaction (Hydrogenation of Oils):
Unsaturated Vegetable Oil+H2→Ni catalyst, ΔSaturated Vegetable Ghee (Vanaspati)\text{Unsaturated Vegetable Oil} + \text{H}_2 \xrightarrow{\text{Ni catalyst, } \Delta} \text{Saturated Vegetable Ghee (Vanaspati)}
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4. Esterification Reaction (Sweet Fruity Smell):
CH3COOH+CH3CH2OH→conc. H2SO4CH3COOCH2CH3 (Ethyl Ethanoate)+H2O\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 \text{ (Ethyl Ethanoate)} + \text{H}_2\text{O}
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5. Saponification (Preparation of Soap):
CH3COOCH2CH3+NaOH⟶CH3COONa (Sodium Ethanoate / Soap)+CH3CH2OH\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \longrightarrow \text{CH}_3\text{COONa} \text{ (Sodium Ethanoate / Soap)} + \text{CH}_3\text{CH}_2\text{OH}
• Micelle Formation & Cleansing Action of Soap
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Soap molecules are sodium/potassium salts of long-chain carboxylic acids (R−COO−Na+R-\text{COO}^-\text{Na}^+).
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Ionic Head (COO−Na+\text{COO}^-\text{Na}^+): Hydrophilic (water-loving), faces outward toward water.
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Hydrocarbon Tail (RR): Hydrophobic (oil-loving), clusters inward, trapping oily dirt droplet at the core of a spherical micelle.
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Scum Formation in Hard Water: In hard water containing Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+} ions, soap forms an insoluble white precipitate called scum. Synthetic detergents (sulfonate salts) do not form scum.
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3. High-Yield Solved Board Examination Questions (3-Mark & 5-Mark)

Solved Board Questions

Standard CBSE board exam questions with complete scientific justifications and marking scheme step protocols.

• 3-Mark Standard Board Question: What is a Homologous Series? State any two of its key characteristics. Give the molecular formula of the 3rd and 4th members of the alkyne series.
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Definition: A homologous series is a family of structurally related organic compounds containing the same functional group, where successive members differ by a −CH2−-\text{CH}_2- unit.
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Key Characteristics:
1. Consecutive members differ by molecular mass of 14 u14\text{ u} (Carbon =12= 12, Hydrogen =2= 2).
2. All members can be represented by a common general molecular formula (e.g. Alkanes: CnH2n+2C_n H_{2n+2}).
3. They exhibit similar chemical properties and a gradual gradation in physical properties (melting point, boiling point, density).
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Alkyne Homologous Series (CnH2n−2C_n H_{2n-2}, where n≥2n \ge 2):
- 1st member (n=2n=2): Ethyne (C2H2\text{C}_2\text{H}_2)
- 2nd member (n=3n=3): Propyne (C3H4\text{C}_3\text{H}_4)
- 3rd member (n=4n=4): Butyne (C4H6\mathbf{\text{C}_4\text{H}_6})
- 4th member (n=5n=5): Pentyne (C5H8\mathbf{\text{C}_5\text{H}_8})
• 5-Mark Comprehensive Question / Numerical: (a) Write balanced chemical equations for the reaction of Ethanoic Acid with: (i) Sodium metal, (ii) Sodium hydroxide, (iii) Sodium hydrogencarbonate.
(b) Explain with a diagram how soap cleans oily clothes in water.
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Part (a) Chemical Reactions of Ethanoic Acid (CH3COOH\text{CH}_3\text{COOH}):
1. With Sodium metal:
2CH3COOH+2Na⟶2CH3COONa (Sodium Ethanoate)+H2(g)↑2\text{CH}_3\text{COOH} + 2\text{Na} \longrightarrow 2\text{CH}_3\text{COONa} \text{ (Sodium Ethanoate)} + \text{H}_2(g)\uparrow
2. With Sodium hydroxide (Neutralization):
CH3COOH+NaOH⟶CH3COONa+H2O(l)\text{CH}_3\text{COOH} + \text{NaOH} \longrightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}(l)
3. With Sodium hydrogencarbonate (Effervescence of CO2\text{CO}_2):
CH3COOH+NaHCO3⟶CH3COONa+H2O(l)+CO2(g)↑\text{CH}_3\text{COOH} + \text{NaHCO}_3 \longrightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}(l) + \text{CO}_2(g)\uparrow
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Part (b) Cleansing Action Mechanism & Micelle Formation:
- Soap molecules have two distinct ends: a hydrophilic polar ionic head (−COO−Na+-\text{COO}^-\text{Na}^+) and a hydrophobic non-polar hydrocarbon tail.
- When soap is dissolved in water around oily cloth, the hydrophobic tails dissolve into the oil droplet while the ionic heads remain in the water phase.
- This forms a radial spherical cluster called a micelle, with oily dirt trapped securely inside the core.
- The ionic heads on the surface repel each other due to like negative charges, preventing micelles from coalescing.
- Mechanical agitation rinses away the emulsified micelles with water, leaving the fabric completely clean.
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4. Practical Laboratory & Competency-Based Case Drill: Esterification & Scented Ester Synthesis (NCERT Activity 4.8)

Practical & Case Drill

Experimental observation analysis, chemical gas tests, and assertion-reason drills.

• Laboratory Activity Context: Esterification & Scented Ester Synthesis (NCERT Activity 4.8)
In a test tube, 1 mL of absolute ethanol is mixed with 1 mL of glacial acetic acid along with a few drops of concentrated sulphuric acid, and heated in a warm water bath.
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Q1: What characteristic smell is detected upon pouring into a beaker of water? →\rightarrow A distinct, pleasant sweet fruity fragrance of ethyl ethanoate ester.
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Q2: What is the role of concentrated sulphuric acid in this reaction? →\rightarrow It acts as a catalyst and dehydrating agent that removes water molecules to drive the equilibrium forward.
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Q3: Write the balanced chemical equation for the reaction. →\rightarrow CH3COOH+C2H5OH→conc. H2SO4CH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \mathbf{\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}}.
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5. CBSE Examiner Marking Scheme, Scientific Notation & Deduction Traps

Important Solved Board Questions

Examiner step-marking allocations, mandatory scientific terminology, and common error avoidance.

• Step-by-Step Marking Rubric & Key Terminology
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1 Mark: Explaining tetravalency and catenation as the root causes of carbon versatility.
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1 Mark: Drawing correct electron dot structures of methane, ethene, and ethanoic acid.
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2 Marks: Writing balanced chemical equations for esterification and saponification.
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1 Mark: Describing micelle structure with hydrophilic head and hydrophobic tail.
• Common Error Deduction Traps
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Trap 1: Forgetting that the first member of the alkyne series is Ethyne (n=2n=2, not n=1n=1).
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Trap 2: Confusing alkaline KMnO4\text{KMnO}_4 (oxidizing agent) with concentrated H2SO4\text{H}_2\text{SO}_4 (dehydrating agent).
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Trap 3: Drawing single bonds instead of double bonds in ethene (C2H4\text{C}_2\text{H}_4) or triple bonds in ethyne (C2H2\text{C}_2\text{H}_2).
Authentic Board Question (3 Marks)Topic: Carbon and its Compounds Chemical Equations & Reaction Mechanisms
Explain the fundamental chemical principles, write the balanced chemical equations with state symbols, and describe the observable phenomena for Carbon and its Compounds.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Balanced Chemical Equation with State Symbols: Write the stoichiometrically balanced chemical equation including state symbols (s,l,g,aq)(s, l, g, aq) for all reactants and products.
1 Mark
Step 2: Observable Physical Changes & Test Confirmation: State key observable indicators: color changes, gas evolution with confirmation test (pop sound/lime water), precipitate formation, or temperature changes.
1 Mark
Step 3: Chemical Reasoning & Mechanism: Explain the underlying chemical mechanism (neutralisation, displacement, redox electron transfer, or functional group properties).
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Carbon and its Compounds in CBSE Science (Chemistry):

1. Balanced Chemical Equation: State the exact equation with mandatory physical states:
Reactant1(s)+Reactant2(aq)⟶Product1(aq)+Product2(g)\text{Reactant}_1(s) + \text{Reactant}_2(aq) \longrightarrow \text{Product}_1(aq) + \text{Product}_2(g)
2. Observable Phenomena: State the exact visual observation (e.g., brisk effervescence, blue to green color shift, white precipitate).
3. Chemical Logic: Explain the underlying ionic dissociation, reactivity series order, or pH shift that drives the reaction.
Examiner Mark Deduction Traps:
•Always include physical state symbols (s),(l),(g),(aq)(s), (l), (g), (aq) in chemical equations to prevent deduction of ½ mark.
•Name specific indicator color changes (e.g. blue litmus to red, pink phenolphthalein turning colorless).

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
Carbon has atomic number 6 with electronic configuration (2,4)(2, 4).
- Gaining 4 electrons to form C4−\text{C}^{4-} would require a nucleus with 6 protons to hold 10 electrons, which is energetically unstable.
- Losing 4 electrons to form C4+\text{C}^{4+} requires an enormous amount of ionization energy to remove 4 electrons from a tiny atom.
Therefore, carbon attains noble gas configuration by sharing electrons (covalent bonding).

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