Chapter 4

Carbon and its Compounds

NCERT solutions and explanations for Class 10 Science Chapter 4 Carbon and its Compounds — covering covalent bonding and electron sharing, carbon tetravalency and catenation, saturated and unsaturated carbon compounds, chains branches and rings, structural isomers, functional groups, homologous series, IUPAC nomenclature basics, combustion of carbon compounds, oxidation of alcohols, addition and substitution reactions, properties and reactions of ethanol, properties and reactions of ethanoic acid, esterification and saponification, soaps and detergents, micelle formation and cleaning action, and hard water and scum.

Questions

16
Q1

Why does carbon not form ionic bonds like sodium or chlorine? Explain why carbon overcomes this limitation by forming covalent bonds instead.

Carbon (atomic number 6) has 4 valence electrons and needs 4 more to attain a stable noble gas configuration. Forming an ionic bond would require either: (i) gaining 4 electrons to become C4\text{C}^{4-}, but the nucleus with only 6 protons cannot hold 10 electrons strongly; or (ii) losing 4 electrons to become C4+\text{C}^{4+}, which requires a very large amount of energy and leaves only 2 electrons for 6 protons to hold. Both options are energetically unfavourable. Carbon overcomes this by sharing its 4 valence electrons with other atoms, forming covalent bonds where both atoms attain a stable outermost shell.
Q2

Draw the electron dot structures for the following molecules and indicate the type of covalent bond (single, double, or triple) formed in each: (a) hydrogen molecule (H2\text{H}_2), (b) oxygen molecule (O2\text{O}_2), (c) nitrogen molecule (N2\text{N}_2), and (d) carbon dioxide (CO2\text{CO}_2).

(a) H2\text{H}_2: Two H atoms share one electron pair → single covalent bond (H ⁣ ⁣H\text{H} \!-\! \text{H}). (b) O2\text{O}_2: Each O atom (atomic number 8, 6 valence electrons) shares two electron pairs with the other → double covalent bond (O ⁣= ⁣O\text{O} \!=\! \text{O}). (c) N2\text{N}_2: Each N atom (atomic number 7, 5 valence electrons) shares three electron pairs → triple covalent bond (N ⁣ ⁣N\text{N} \!\equiv\! \text{N}). (d) CO2\text{CO}_2: Carbon shares two electron pairs with each of the two oxygen atoms → two double bonds (O ⁣= ⁣C ⁣= ⁣O\text{O} \!=\! \text{C} \!=\! \text{O}).
Q3

What are the two properties of carbon that lead to the enormous variety of carbon compounds? Explain each property with suitable examples.

The two properties are (i) catenation — the ability of carbon atoms to form bonds with other carbon atoms, producing long chains, branched chains, and rings; and (ii) tetravalency — carbon has a valency of 4, enabling it to bond with four other atoms, including other carbon atoms and monovalent elements such as hydrogen, chlorine, and elements like oxygen, nitrogen, and sulphur. Catenation: carbon atoms can link via single, double, or triple bonds to form compounds with very large numbers of carbon atoms. Silicon also forms chains but only up to 7–8 atoms because Si–Si bonds are weaker. Tetravalency: carbon's small atomic size allows its nucleus to hold shared electron pairs strongly, forming stable bonds with many elements. Together, these properties explain why millions of carbon compounds exist.
Q4

Describe the three allotropes of carbon — diamond, graphite, and buckminsterfullerene — with their structures and one key property of each. Why do they have different physical properties despite being made of the same element?

The three allotropes are different structural forms of elemental carbon with the same chemical properties but different physical properties. (i) Diamond: each carbon atom is bonded to four other carbon atoms in a rigid 3D tetrahedral network. It is the hardest known natural substance. (ii) Graphite: each carbon atom is bonded to three others in a hexagonal planar array, forming layers stacked one above the other. It is soft, slippery, and a good conductor of electricity (unlike most non-metals). (iii) Buckminsterfullerene (C60\text{C}_{60}): carbon atoms arranged in a football-like spherical structure, named after architect Buckminster Fuller. Their physical properties differ because the arrangement and bonding of carbon atoms are different in each allotrope.
Q5

Distinguish between saturated and unsaturated carbon compounds with the help of examples — ethane (C2H6\text{C}_2\text{H}_6), ethene (C2H4\text{C}_2\text{H}_4), and ethyne (C2H2\text{C}_2\text{H}_2). Draw their electron dot structures. Which type is more reactive and why?

Saturated compounds contain only single bonds between carbon atoms. Example: ethane (C2H6\text{C}_2\text{H}_6) — each carbon is linked by a single C ⁣ ⁣C\text{C} \!-\! \text{C} bond with remaining valencies satisfied by hydrogen. Electron dot structure shows one shared electron pair between carbon atoms and one shared pair between each carbon and three hydrogens. Unsaturated compounds contain double or triple bonds. Examples: ethene (C2H4\text{C}_2\text{H}_4) has a C ⁣= ⁣C\text{C} \!=\! \text{C} double bond; ethyne (C2H2\text{C}_2\text{H}_2) has a C ⁣ ⁣C\text{C} \!\equiv\! \text{C} triple bond. Unsaturated compounds are more reactive because the double or triple bonds can break and add atoms without losing carbon atoms — the π\pi bonds in the multiple bonds are weaker and more accessible to attacking reagents.
Q6

What are structural isomers? Using the molecular formula C4H10\text{C}_4\text{H}_{10}, draw the two possible structural isomers. How do chains, branches, and rings expand the diversity of carbon compounds?

Structural isomers are compounds with the same molecular formula but different structures (different arrangement of atoms). For C4H10\text{C}_4\text{H}_{10}, the two isomers are: (i) n-butane — a straight chain of four carbon atoms (CH3 ⁣ ⁣CH2 ⁣ ⁣CH2 ⁣ ⁣CH3\text{CH}_3 \!-\! \text{CH}_2 \!-\! \text{CH}_2 \!-\! \text{CH}_3), and (ii) isobutane — a branched chain with three carbons in a row and one methyl (CH3\text{CH}_3) branch on the central carbon. Carbon chains can be straight (n-butane), branched (isobutane), or cyclic (cyclohexane C6H12\text{C}_6\text{H}_{12}, benzene C6H6\text{C}_6\text{H}_6). These three modes — chains, branches, and rings — combined with single, double, and triple bonds and the possibility of various functional groups — generate the enormous diversity of organic compounds.
Q7

What is a functional group? List the main functional groups in organic chemistry, giving the formula and the class of compounds for each. Explain why compounds with the same functional group show similar chemical properties.

A functional group is a heteroatom (such as Cl, Br, O, N, S) or a group of atoms containing heteroatoms that, when attached to a carbon chain, confers specific chemical properties to the compound, regardless of the chain length. Main functional groups: (i) Halo ( ⁣ ⁣Cl\!-\!\text{Cl} /  ⁣ ⁣Br\!-\!\text{Br}) → haloalkanes; (ii) Alcohol ( ⁣ ⁣OH\!-\!\text{OH}); (iii) Aldehyde ( ⁣ ⁣CHO\!-\!\text{CHO}); (iv) Ketone ( ⁣ ⁣CO ⁣ ⁣\!-\!\text{CO} \!-\!); (v) Carboxylic acid ( ⁣ ⁣COOH\!-\!\text{COOH}). Compounds with the same functional group show similar chemical properties because the chemical reactivity is determined primarily by the functional group, not the carbon chain length.
Q8

What is a homologous series? Explain using the example of alcohols from methanol to butanol. How do melting points, boiling points, and chemical properties change as we move up a homologous series?

A homologous series is a series of carbon compounds in which the same functional group substitutes for hydrogen in a carbon chain, and successive members differ by a  ⁣ ⁣CH2 ⁣ ⁣\!-\!\text{CH}_2 \!-\! unit. The alcohol series: methanol (CH3OH\text{CH}_3\text{OH}), ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}), propanol (C3H7OH\text{C}_3\text{H}_7\text{OH}), butanol (C4H9OH\text{C}_4\text{H}_9\text{OH}). Each successive member differs by one  ⁣ ⁣CH2 ⁣ ⁣\!-\!\text{CH}_2 \!-\! group (molecular mass difference of 14 u14\text{ u}). Physical properties: melting and boiling points increase with increasing molecular mass due to stronger intermolecular forces. Solubility in a given solvent shows a similar gradation. Chemical properties remain very similar because they are determined by the same functional group ( ⁣ ⁣OH\!-\!\text{OH} in this case).
Q9

Explain the system of naming (nomenclature) of carbon compounds. Using this system, name the following compounds: (a) CH3 ⁣ ⁣CH2 ⁣ ⁣Br\text{CH}_3 \!-\! \text{CH}_2 \!-\! \text{Br}, (b) CH3 ⁣ ⁣CH2 ⁣ ⁣COOH\text{CH}_3 \!-\! \text{CH}_2 \!-\! \text{COOH}, and (c) CH3 ⁣ ⁣CO ⁣ ⁣CH3\text{CH}_3 \!-\! \text{CO} \!-\! \text{CH}_3.

Nomenclature follows these rules: (1) Identify the number of carbon atoms in the longest chain — meth(1), eth(2), prop(3), but(4), etc. (2) Identify the functional group and use its appropriate prefix or suffix: halo (prefix chloro/bromo), alcohol (suffix -ol), aldehyde (suffix -al), ketone (suffix -one), carboxylic acid (suffix -oic acid), alkenes (suffix -ene), alkynes (suffix -yne). (3) If the suffix begins with a vowel, drop the final 'e' of the carbon chain name. (4) For unsaturation, replace 'ane' with 'ene' or 'yne'. Naming: (a) CH3 ⁣ ⁣CH2 ⁣ ⁣Br\text{CH}_3 \!-\! \text{CH}_2 \!-\! \text{Br} → two carbons → ethane → prefix 'bromo' → bromoethane. (b) CH3 ⁣ ⁣CH2 ⁣ ⁣COOH\text{CH}_3 \!-\! \text{CH}_2 \!-\! \text{COOH} → three carbons → propane → suffix -oic acid (drop 'e') → propanoic acid. (c) CH3 ⁣ ⁣CO ⁣ ⁣CH3\text{CH}_3 \!-\! \text{CO} \!-\! \text{CH}_3 → three carbons → propane → suffix -one (drop 'e') → propanone.
Q10

Describe the combustion of carbon compounds. Why do saturated hydrocarbons generally burn with a clean blue flame while unsaturated hydrocarbons burn with a yellow sooty flame? Write balanced equations for the combustion of methane and ethanol.

Carbon compounds burn in oxygen to release carbon dioxide, water, heat, and light. Saturated hydrocarbons (alkanes) burn with a clean blue flame because the carbon-to-hydrogen ratio is low — complete combustion is achieved and all carbon oxidises to CO2\text{CO}_2. Unsaturated hydrocarbons have a higher carbon proportion — they undergo incomplete combustion, producing unburnt carbon particles (soot) that glow yellow when heated. The blackened bottom of a cooking vessel indicates blocked air holes and incomplete combustion. Balanced equations: CH4+2O2CO2+2H2O+heat and light\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{heat and light}; CH3CH2OH+3O22CO2+3H2O+heat and light\text{CH}_3\text{CH}_2\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{heat and light}.
Q11

How is ethanol converted to ethanoic acid? Why is this reaction classified as an oxidation reaction? Name the oxidising agents used and explain the observation that indicates the reaction is over.

Ethanol (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}) is oxidised to ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) using alkaline potassium permanganate (KMnO4\text{KMnO}_4) or acidified potassium dichromate (K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7) as oxidising agents, with heating: CH3CH2OHor acidified K2Cr2O7+HeatAlkaline KMnO4+HeatCH3COOH\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{or acidified }\text{K}_2\text{Cr}_2\text{O}_7 + \text{Heat}]{\text{Alkaline }\text{KMnO}_4 + \text{Heat}} \text{CH}_3\text{COOH}. This is an oxidation reaction because ethanol gains oxygen — the oxygen atom from the oxidising agent adds to the ethanol molecule. During the activity, the purple colour of KMnO4\text{KMnO}_4 disappears initially as it oxidises ethanol. When excess KMnO4\text{KMnO}_4 is added and the purple colour persists, it indicates that all ethanol has been oxidised and the reaction is complete.
Q12

Distinguish between addition and substitution reactions with one chemical equation each. What is hydrogenation and why is it industrially important?

Addition reactions occur in unsaturated hydrocarbons where atoms (e.g., hydrogen) add across the double or triple bond to form a saturated compound, catalysed by nickel or palladium. Example: hydrogenation of ethene — CH2 ⁣= ⁣CH2+H2Ni catalystCH3 ⁣ ⁣CH3\text{CH}_2 \!=\! \text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{CH}_3 \!-\! \text{CH}_3. Substitution reactions occur in saturated hydrocarbons where an atom (e.g., chlorine) replaces a hydrogen atom, typically requiring sunlight. Example: CH4+Cl2SunlightCH3Cl+HCl\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{Sunlight}} \text{CH}_3\text{Cl} + \text{HCl}. Hydrogenation is an addition reaction that converts unsaturated compounds to saturated ones by adding hydrogen. Industrially, it converts liquid vegetable oils (unsaturated) into solid vanaspati ghee (saturated fats) using a nickel catalyst. Vegetable oils with long unsaturated carbon chains are healthier for cooking; animal fats with saturated fatty acids are considered harmful.
Q13

Describe two important chemical reactions of ethanol. Explain how ethanol reacts with sodium and how it can be converted to ethene. What is the role of concentrated sulphuric acid in the dehydration reaction?

Two key reactions of ethanol: (1) Reaction with sodium: Ethanol reacts with sodium metal to produce sodium ethoxide and hydrogen gas: 2Na+2CH3CH2OH2CH3CH2ONa++H22\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2 \uparrow. The gas can be tested by the pop sound with a burning splinter (characteristic of hydrogen). (2) Dehydration to ethene: When ethanol is heated at 443 K443\text{ K} with excess concentrated sulphuric acid, it undergoes dehydration — a water molecule is removed, forming ethene: CH3 ⁣ ⁣CH2OHH2SO4Hot Conc.CH2 ⁣= ⁣CH2+H2O\text{CH}_3\!-\!\text{CH}_2\text{OH} \xrightarrow[\text{H}_2\text{SO}_4]{\text{Hot Conc.}} \text{CH}_2\!=\!\text{CH}_2 + \text{H}_2\text{O}. Concentrated H2SO4\text{H}_2\text{SO}_4 acts as a dehydrating agent — it removes the elements of water from ethanol, converting the saturated alcohol into an unsaturated hydrocarbon.
Q14

Describe the esterification reaction between ethanoic acid and ethanol. Write the chemical equation, list the conditions required, and give two practical uses of esters. What is saponification and how is it related to soap making?

Esterification is the reaction of a carboxylic acid with an alcohol in the presence of an acid catalyst (concentrated H2SO4\text{H}_2\text{SO}_4) to form an ester and water: CH3COOH+C2H5OHConc. H2SO4CH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}. The mixture is warmed in a water bath for about 5 minutes and then poured into water — a sweet, fruity smell indicates ester formation. Esters are used in making perfumes and as flavouring agents. Saponification is the reverse — treating an ester with an alkali (NaOH\text{NaOH}) to regenerate the alcohol and the sodium salt of the carboxylic acid: CH3COOC2H5NaOHC2H5OH+CH3COONa\text{CH}_3\text{COOC}_2\text{H}_5 \xrightarrow{\text{NaOH}} \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}. Soaps are sodium or potassium salts of long-chain carboxylic acids, prepared by the alkaline hydrolysis of esters (fats and oils).
Q15

What are soaps? Explain the mechanism of the cleaning action of soap with the help of micelle formation. Why does a soap solution appear cloudy?

Soaps are sodium (Na+\text{Na}^+) or potassium (K+\text{K}^+) salts of long-chain carboxylic acids (fatty acids). A soap molecule has a dual nature: a long hydrophobic hydrocarbon tail (non-polar, water-hating, oil-loving) and a hydrophilic ionic head (COONa+\text{COO}^-\text{Na}^+, polar, water-loving). When soap is dissolved in water, the hydrophobic tails cluster together inward, away from water, while the hydrophilic heads face outward — forming spherical aggregates called micelles. Oily/greasy dirt gets trapped in the hydrophobic core of the micelle. The micelle, with the trapped dirt, remains suspended as a colloid (due to ion–ion repulsion between negatively charged heads), and is easily rinsed away with water. Soap solutions appear cloudy because micelles are large enough to scatter light.
Q16

Why does soap not work effectively in hard water? What is scum and how is it formed? How do detergents overcome this limitation? Explain the difference in chemical composition between soaps and detergents.

Hard water contains dissolved calcium (Ca2+\text{Ca}^{2+}) and magnesium (Mg2+\text{Mg}^{2+}) salts (hydrogencarbonates, sulphates, or chlorides). Soap reacts with these ions to form an insoluble, white, curdy precipitate called scum (calcium or magnesium salts of fatty acids). This scum wastes soap, prevents foam formation, and sticks to fabrics. Soap only works after all hardness ions are precipitated. Detergents overcome this because they are generally sodium salts of sulphonic acids or ammonium salts, and their charged ends do NOT form insoluble precipitates with Ca2+\text{Ca}^{2+} or Mg2+\text{Mg}^{2+} ions. Thus, detergents remain effective cleansing agents even in hard water. Composition difference: soaps are sodium/potassium salts of long-chain carboxylic acids (RCOONa+\text{RCOO}^-\text{Na}^+); detergents are sodium salts of long-chain sulphonic acids (RSO3Na+\text{RSO}_3^-\text{Na}^+) or similar compounds with different anionic groups.