Class 10 Science · Chapter 4

Carbon and its Compounds

442 Class 10 Science questions on Carbon and its Compounds from the NCERT textbook, NCERT Exemplar and CBSE board papers 2017–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 7 of 23

1. Q121 CBSE 2026
   1. (i)State any two reasons for carbon forming a large number of compounds. ​Why ​does carbon ​form compounds mainly by covalent bonds ?
   2. (ii)Identify ​the ​compound which contains ​aldehyde as its ​functional group : ${\text{C}}_3{\text{H}}_7\text{OH}$, ${\text{C}}_3{\text{H}}_7\text{Cl}$, ${\text{C}}_2{\text{H}}_5\text{CHO}$, ${\text{CH}}_3{\text{COCH}}_3$
   Show answer

   Catenation and tetravalency let carbon form many compounds; carbon shares its four valence electrons, forming covalent bonds; ${\text{C}}_2{\text{H}}_5\text{CHO}$ has the aldehyde group.

   (i) Two reasons: (1) Catenation — carbon atoms link to one another by strong covalent bonds and build long straight chains, branched chains and closed rings; (2) Tetravalency — carbon has four valence electrons, so one carbon atom can bond with four other atoms of carbon or of other monovalent elements, and because the carbon atom is small these bonds are strong and stable. Carbon forms compounds mainly by covalent bonds because it has four electrons in its outermost shell: it can neither gain four electrons (a nucleus of six protons cannot hold ten electrons) nor lose four electrons (that would need a very large amount of energy), so it attains a noble-gas arrangement by **sharing** its four valence electrons, and a shared pair of electrons is a covalent bond. (ii) ${\text{C}}_2{\text{H}}_5\text{CHO}$ (propanal) — it carries the aldehyde group $-\text{CHO}$.

   Step-by-step solution →
2. Q122 CBSE 2026
   Give reasons for ​the following :

   1. (i)Unsaturated ​hydrocarbons show ​addition reactions.
   2. (ii)Detergents are ​better cleansing agents than soaps.
   3. (iii)Carbon ​forms ​only covalent compounds.
   Show answer

   Multiple bonds open to add atoms; detergents give no scum in hard water; carbon shares electrons as gaining or losing four is difficult.

   (i) Unsaturated hydrocarbons contain a carbon-carbon double or triple bond, so their carbon atoms are not bonded to the maximum number of atoms. One part of the multiple bond opens and the new atoms add on with nothing expelled, as in ${\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3{\text{CH}}_3$; (ii) Soaps form insoluble calcium and magnesium salts - a scum - with the ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ ions of hard water, so they are wasted and do not lather, whereas the charged ends of detergents form no such precipitate and go on cleaning even in hard water; (iii) Carbon has four electrons in its outermost shell, and it can neither gain four electrons (a nucleus of six protons cannot hold ten electrons) nor lose four (far too much energy is needed), so it completes its shell by sharing its four electrons and forms only covalent bonds.

   Step-by-step solution →
3. Q123 CBSE 2026
   An ​organic compound ​‘X’ is a ​constituent of wine. ‘X’ on reacting with ​acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ ​forms another compound ‘Y’ with molecular formula ${\text{C}}_2{\text{H}}_4{\text{O}}_2$. ‘Y’ produces brisk ​effervescence with sodium carbonate.

   1. (i)Identify ‘X’ and ‘Y’.
   2. (ii)Name the gas ​evolved when ‘Y’ reacts with sodium carbonate.
   3. (iii)Write the chemical equations involved for the chemical ​reactions taking place.
   Show answer

   X is ethanol and Y is ethanoic acid; the gas evolved is carbon dioxide.

   (i) ‘X’ is ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$ — the alcohol present in wine; ‘Y’ is ethanoic acid (acetic acid), ${\text{CH}}_3\text{COOH}$, whose molecular formula is ${\text{C}}_2{\text{H}}_4{\text{O}}_2$. (ii) The gas evolved is carbon dioxide, ${\text{CO}}_2$. (iii) ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{\text{Heat}}{\overset{{\text{acidified K}}_2{\text{Cr}}_2{\text{O}}_7}{\to }}{\text{CH}}_3\text{COOH}$ and $2{\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3⟶2{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$

   Step-by-step solution →
4. Q124 CBSE 2026
   Alcohol forms ​a homologous series with general formula ${\text{C}}_n{\text{H}}_{2n+1}\text{–OH}$ and $\text{–OH}$ group as ​functional group. Ethanol is ​commonly called alcohol and ​is ​used in ​alcoholic drinks. It is good solvent, used ​in ​medicines, cough syrups, tonics etc.

   1. (a)Write structural formula and name of 4th member of alcohol homologous series.[1]
   2. (b)What happens when ethanol is heated with alkaline ${\text{KMnO}}_4$ ? Write chemical equation involved.[1]
   3. (c)Write the chemical equation of reaction of ethanol with ethanoic acid in the presence of concentrated ${\text{H}}_2{\text{SO}}_4$. Write the name of this reaction.[2]
   4. (c)OR : What happens when ethanol is heated with excess concentrated sulphuric acid at $443$ K ? Write chemical equation involved. What is the role of concentrated sulphuric acid in this reaction ?[2]
   Show answer

   The 4th alcohol is butanol, ${\text{C}}_4{\text{H}}_9\text{OH}$; ethanol is oxidised to ethanoic acid; ethanol and ethanoic acid give an ester by esterification.

   (a) The 4th member is butanol (butan-1-ol), ${\text{C}}_4{\text{H}}_9\text{OH}$; its structural formula is ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$, drawn out in full as $$\begin{array}{ccccccccccccc}& & \text{H}& & \text{H}& & \text{H}& & \text{H}& & & & \\ & & |& & |& & |& & |& & & & \\ \text{H}& -& \text{C}& -& \text{C}& -& \text{C}& -& \text{C}& -& \text{O}& -& \text{H}\\ & & |& & |& & |& & |& & & & \\ & & \text{H}& & \text{H}& & \text{H}& & \text{H}& & & & \end{array}$$ (b) Ethanol is oxidised to ethanoic acid (the purple colour of the permanganate fades): ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\underset{\text{heat}}{\overset{{\text{alkaline KMnO}}_4}{\to }}{\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$. (c) ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$; the reaction is esterification and the product, ethyl ethanoate, is a sweet-smelling ester. OR (c) Ethanol is dehydrated to ethene: ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{hot conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$; the concentrated sulphuric acid acts as a dehydrating agent.

   Step-by-step solution →
5. Q125 CBSE 2026
   1. (i)Carbon cannot form ${\mathrm{C}}^{4+}$ cation and ${\mathrm{C}}^{4-}$ anion. Explain.
   2. (ii)​Define homologous series of carbon compounds. Why are the melting and ​boiling points of ${\mathrm{C}}_4{\mathrm{H}}_{10}$ higher than ${\mathrm{C}}_3{\mathrm{H}}_8$ or ${\mathrm{C}}_2{\mathrm{H}}_6$ ?
   3. (iii)​What ​happens when ethanol ​is heated with acidified potassium dichromate ?
   Show answer

   Carbon cannot lose or gain four electrons, so it shares them; heavier ${\text{C}}_4{\text{H}}_{10}$ has stronger intermolecular forces; acidified dichromate oxidises ethanol to ethanoic acid.

   (i) Carbon has four valence electrons (configuration $2,4$). To form ${\mathrm{C}}^{4+}$ it would have to lose all four, which needs an enormous amount of energy and would leave a nucleus of six protons holding only two electrons; to form ${\mathrm{C}}^{4-}$ it would have to gain four, and a nucleus of only six protons cannot hold on to ten electrons. Carbon therefore shares its four valence electrons instead, forming four covalent bonds - it is tetravalent. (ii) A homologous series is a series of carbon compounds having the same general formula and the same functional group, in which successive members differ by one $-\mathrm{C}{\mathrm{H}}_2-$ unit, that is by 14 u; the members have similar chemical properties and a gradual gradation in physical properties. ${\mathrm{C}}_4{\mathrm{H}}_{10}$ (58 u) is a larger, heavier molecule than ${\mathrm{C}}_3{\mathrm{H}}_8$ (44 u) or ${\mathrm{C}}_2{\mathrm{H}}_6$ (30 u), so the intermolecular forces of attraction between its molecules are stronger and more energy is needed to separate them; hence its higher melting and boiling points. (iii) Acidified potassium dichromate acts as an oxidising agent and oxidises ethanol to ethanoic acid, the orange colour of the dichromate turning green: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{acidified }{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7,\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   Step-by-step solution →
6. Q126 CBSE 2026
   1. (i)Give reasons for the following : (I) Covalent ​compounds are ​poor conductor ​of electricity. (II) Soap does not ​form lather ​in ​hard water. (III) Carbon ​shows catenation but silicon does not.
   2. (ii)Write chemical equations for the following : (I) Oxidation of ethanol by acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$. (II) Hydrogenation of ​ethene.
   Show answer

   No free ions in covalent compounds; scum forms in hard water; strong C-C bonds give catenation; ethanol gives ethanoic acid, ethene gives ethane.

   (i) (I) In covalent compounds the atoms share electrons, so no ions are formed and there are no free charged particles to carry a current; hence they are poor conductors of electricity. (II) Hard water contains calcium and magnesium salts. Soap reacts with the ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ ions to form an insoluble precipitate (scum), so the soap is used up and no lather forms. (III) The carbon atom is small, so carbon–carbon bonds are very strong and stable and carbon atoms can join into long chains and rings. Silicon atoms are larger and silicon–silicon bonds are weak, so silicon forms only short, very reactive chains (up to seven or eight atoms). (ii) (I) ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\stackrel{{\text{acidified K}}_2{\text{Cr}}_2{\text{O}}_7,\text{heat}}{\to }{\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$ (ethanol to ethanoic acid). (II) ${\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni catalyst}}{\to }{\text{CH}}_3-{\text{CH}}_3$ (ethene to ethane).

   Step-by-step solution →
7. Q127 CBSE 2026
   A carbon compound ‘P’ of molecular formula ${\text{C}}_2{\text{H}}_6\text{O}$ is found ​to be neutral when tested with red and blue litmus. A gas that burns ​with a pop sound is produced when sodium metal ​reacts with compound ‘P’.

   1. (i)Identify the compound ‘P’.
   2. (ii)The ​carbon compound ‘P’ is heated with conc. ${\text{H}}_2{\text{SO}}_4$ at 443 K to produce compound ‘Q’. (I) Write the chemical equation ​for ​the reaction. (II) What is the role of conc. ${\text{H}}_2{\text{SO}}_4$ in this reaction ?
   3. (iii)What is likely to be observed on heating compound ‘P’ with ethanoic acid with an ​acid as a catalyst ? Write the chemical equation for the reaction.
   Show answer

   P is ethanol; heated with conc. sulphuric acid at 443 K it dehydrates to ethene (Q); with ethanoic acid it forms a sweet-smelling ester.

   (i) ‘P’ is ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$ (${\text{CH}}_3{\text{CH}}_2\text{OH}$). It is neutral to litmus, and sodium releases hydrogen from it, the gas that burns with a pop sound. (ii) (I) ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{hot conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$; ‘Q’ is ethene. (II) Concentrated sulphuric acid acts as a dehydrating agent: it removes water from ethanol. (iii) A sweet-smelling (fruity) substance, the ester ethyl ethanoate, is formed (esterification): ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{\text{acid}}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$

   Step-by-step solution →
8. Q128 CBSE 2026
   1. (i)What is ​homologous ​series ? Which two of the ​following ​organic compounds belong to the ​same homologous series ? ${\text{C}}_2{\text{H}}_4$, ${\text{C}}_4{\text{H}}_{10}$, ${\text{C}}_3{\text{H}}_4$, ${\text{C}}_4{\text{H}}_8$
   2. (ii)Write the product(s) in the ​following reactions : (I) ${\text{CH}}_3-{\text{CH}}_2-\text{OH}+{\text{CH}}_3\text{COOH}\stackrel{\text{Acid}}{\to }$ (II) ${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\stackrel{\text{Burning}}{\to }$
   3. (iii)Why does micelle ​formation take place when soap is added to ​water ?
   Show answer

   ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_4{\text{H}}_8$ are in one series; products are ethyl ethanoate with water, and carbon dioxide with water; soap tails cluster inward, forming micelles.

   (i) A homologous series is a series of carbon compounds having the same functional group and the same general formula, in which successive members differ by a $-{\text{CH}}_2-$ unit ($14$ u); they have similar chemical properties and a gradation in physical properties. ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_4{\text{H}}_8$ belong to the same series (alkenes, ${\text{C}}_n{\text{H}}_{2n}$). (ii) (I) Ethyl ethanoate and water: ${\text{CH}}_3{\text{CH}}_2\text{OH}+{\text{CH}}_3\text{COOH}\stackrel{\text{Acid}}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$. (II) Carbon dioxide and water, with heat and light: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}$. (iii) A soap molecule has an ionic (hydrophilic) end that interacts with water and a long hydrocarbon (hydrophobic) tail that is repelled by water but interacts with oil and dirt. In water the tails cluster together inside, around any oily dirt, while the ionic ends stay outside facing the water, so the soap molecules form spherical clusters called micelles.

   Step-by-step solution →
9. Q129 CBSE 2026
   1. (i)Define catenation. Why ​does ​carbon show catenation ?
   2. (ii)Define ​homologous series. ​Give examples of two homologues ​whose ​molecular mass differ by 14 amu.
   3. (iii)Why do soaps not work well in hard water ​?
   Show answer

   Carbon is tetravalent with strong $\text{C}-\text{C}$ bonds; methane and ethane differ by 14 u; soap forms insoluble scum in hard water.

   (i) Catenation is the property of an element whose atoms link with one another by covalent bonds to form long straight chains, branched chains and closed rings. Carbon shows it because it is tetravalent - four valence electrons, so four covalent bonds per atom - and because the carbon atom is small, which makes the C to C bond short and very strong (about $348$ kJ mol${}^{-1}$), so long chains remain stable. (ii) A homologous series is a series of organic compounds with the same general formula and similar chemical properties in which successive members differ by one $-{\text{CH}}_2-$ group, that is by $14$ u. Example of two such homologues: methane ${\text{CH}}_4$ ($16$ u) and ethane ${\text{C}}_2{\text{H}}_6$ ($30$ u), since $30-16=14$ u. (iii) Hard water contains dissolved calcium and magnesium salts. Their ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ ions react with the soap to form insoluble calcium and magnesium salts of the fatty acid - a curdy white scum - so the soap is used up instead of forming lather and cleans poorly: $2{\text{C}}_{17}{\text{H}}_{35}\text{COONa}+{\text{Ca}}^{2+}⟶({\text{C}}_{17}{\text{H}}_{35}\text{COO}{)}_2\text{Ca}\downarrow +2{\text{Na}}^{+}$.

   Step-by-step solution →
10. Q130 CBSE 2026
    1. (i)Give reason why carbon can neither form ${\text{C}}^{4+}$ cation nor ${\text{C}}^{4-}$ anion, but forms covalent compounds.
    2. (ii)What happens when ​: (I) a piece of ​sodium is ​dropped into ethanol ? (II) ethanol ​is ​heated with ethanoic acid in the presence of an ​acid ?
    3. (iii)Define soap. Why are soaps not effective ​in hard water ​?
    Show answer

    Carbon shares electrons; sodium gives hydrogen and sodium ethoxide; ethanol with ethanoic acid gives an ester; soap forms insoluble scum in hard water.

    (i) Carbon (2, 4) has four valence electrons. It cannot form ${\text{C}}^{4+}$, because removing four electrons needs a very large amount of energy (six protons would be left holding only two electrons), and it cannot form ${\text{C}}^{4-}$, because a nucleus of six protons cannot hold ten electrons (four extra). So carbon shares its four valence electrons with other atoms and forms covalent compounds. (ii) (I) Sodium reacts with ethanol to form sodium ethoxide, and hydrogen gas is evolved: $2\text{Na}+2{\text{CH}}_3{\text{CH}}_2\text{OH}⟶2{\text{CH}}_3{\text{CH}}_2\text{ONa}+{\text{H}}_2$. (II) A sweet-smelling ester, ethyl ethanoate, is formed along with water (esterification): ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{\text{acid}}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$. (iii) Soap is the sodium or potassium salt of a long-chain carboxylic acid (e.g. sodium stearate, ${\text{C}}_{17}{\text{H}}_{35}\text{COONa}$). In hard water it reacts with the calcium and magnesium ions to form an insoluble scum, so the soap is wasted and does not lather or clean well.

    Step-by-step solution →
11. Q131 CBSE 2026
    1. (i)How ​would you name the following : I. $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}$ II. $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$
    2. (ii)When ethanol reacts ​with ethanoic acid ​in the ​presence of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, a ​substance with sweet smell is produced. ​Answer the following : I. State the class of compounds to which the sweet smelling ​compounds belong. II. Write the chemical equation for the ​reaction.
    3. (iii)Why soaps do not produce foam with hard water ?

    Show answer

    Propanal and propyne; the sweet-smelling compounds are esters; soap forms insoluble scum with the calcium and magnesium ions of hard water, so no foam.

    (i) I. Propanal ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$), an aldehyde. II. Propyne ($\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$), also written prop-1-yne. (ii) I. The sweet smelling compounds are esters, and the reaction that forms them is called esterification. II. $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\text{conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$, that is ethanoic acid and ethanol give ethyl ethanoate and water. (iii) Hard water contains $\mathrm{C}{\mathrm{a}}^{2+}$ and $\mathrm{M}{\mathrm{g}}^{2+}$ ions. These react with soap, which is the sodium salt of a long chain carboxylic acid, to form insoluble calcium and magnesium salts of that acid - a curdy white precipitate called scum. The soap is used up as scum instead of dissolving to form micelles, so it does not lather and a large amount of soap is wasted.

    Step-by-step solution →
12. Q132 CBSE 2026
    Mohan heated ​ethanol with a compound ‘X’ in the presence of a few drops of ​conc. ${\text{H}}_2{\text{SO}}_4$ and ​observed a sweet smelling compound ‘Y’ is formed. When ‘Y’ ​is treated with sodium ​hydroxide it gives back ​ethanol and a compound ‘Z’.

    1. (i)Identify ‘X’, ‘Y’ and ‘Z’.
    2. (ii)Write the role of conc. ${\text{H}}_2{\text{SO}}_4$ in the reaction.
    3. (iii)​Write the chemical equations involved and name the reactions.
    Show answer

    X is ethanoic acid, Y is ethyl ethanoate and Z is sodium ethanoate; concentrated sulphuric acid is a catalyst and a dehydrating agent.

    (i) X is ethanoic acid (acetic acid), ${\text{CH}}_3\text{COOH}$; Y is ethyl ethanoate (ethyl acetate), ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5$, a sweet-smelling ester; Z is sodium ethanoate (sodium acetate), ${\text{CH}}_3\text{COONa}$. (ii) Concentrated ${\text{H}}_2{\text{SO}}_4$ acts as a catalyst for the esterification; as a dehydrating agent it also takes up the water formed. (iii) Esterification: ${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$. Saponification: ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}\to {\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3\text{COONa}$.

    Step-by-step solution →
13. Q133 CBSE 2026
    1. (i)Write the ​structural formula of ​benzene.
    2. (ii)State the reason why carbon can ​neither ​form ${\text{C}}^{4+}$ cation nor ${\text{C}}^{4-}$ anion.
    3. (iii)What type of bond is formed between carbon and ​hydrogen ?
    4. (iv)How is ethanol converted ​to ​ethanoic acid ?
    5. (v)Complete the following chemical equation : ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶$
    Show answer

    Benzene is a hexagonal ring of six carbons; carbon forms covalent bonds; ethanol is oxidised to ethanoic acid; the ester gives sodium ethanoate and ethanol.

    (i) Benzene, ${\text{C}}_6{\text{H}}_6$: a hexagonal ring of six carbon atoms joined by alternate single and double bonds, with one hydrogen atom joined to each carbon (drawn in the figure at step 1). (ii) Carbon has four valence electrons (configuration $2,4$). To form ${\text{C}}^{4-}$ it would have to gain four electrons, and its nucleus of six protons cannot easily hold ten electrons; to form ${\text{C}}^{4+}$ it would have to lose four electrons, which needs a very large amount of energy. So carbon shares its valence electrons and forms covalent bonds instead. (iii) A covalent bond: a single covalent bond, formed by sharing one pair of electrons between carbon and hydrogen. (iv) By oxidation: ethanol is heated with alkaline potassium permanganate (or acidified potassium dichromate): ${\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{alkaline KMnO}}_4+\text{heat}}{\to }{\text{CH}}_3\text{COOH}$ (v) ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$ (sodium ethanoate and ethanol; the reaction is saponification).

    Step-by-step solution →
14. Q134 CBSE 2026
    1. (i)Name ​the following compounds : (I) ${\text{CH}}_3-\underset{\underset{\text{O}}{\Vert }}{C}-{\text{CH}}_3$ (II) $\text{H}-\stackrel{\stackrel{\text{H}}{|}}{\underset{\underset{\text{H}}{|}}{C}}-\text{C}\equiv \text{C}-\text{H}$
    2. (ii)Name the reaction which is commonly used in ​the conversion of vegetable oils to fats.
    3. (iii)Why is the melting point of ${\text{C}}_4{\text{H}}_{10}$ higher ​than ${\text{C}}_3{\text{H}}_8$ ​?
    4. (iv)What ​happens when a ​piece of sodium is dropped into ​ethanol ?
    5. (v)Write ​the products in the following reaction : ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶$
    Show answer

    Propanone and propyne; hydrogenation; ${\text{C}}_4{\text{H}}_{10}$ has the higher molecular mass; sodium gives hydrogen and sodium ethoxide; the ester gives sodium ethanoate and ethanol.

    (i) (I) Propanone (acetone); (II) Propyne. (ii) Hydrogenation: hydrogen is added to the vegetable oil in the presence of a nickel (or palladium) catalyst. (iii) ${\text{C}}_4{\text{H}}_{10}$ has one more $-{\text{CH}}_2-$ unit and a higher molecular mass ($58$ u) than ${\text{C}}_3{\text{H}}_8$ ($44$ u); in a homologous series melting points rise with molecular mass, as the larger molecules attract one another more strongly. (iv) Sodium reacts with ethanol, giving off bubbles of hydrogen gas and forming sodium ethoxide: $2\text{Na}+2{\text{CH}}_3{\text{CH}}_2\text{OH}\to 2{\text{CH}}_3{\text{CH}}_2\text{ONa}+{\text{H}}_2$. (v) Sodium ethanoate and ethanol: ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}\to {\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$ (saponification).

    Step-by-step solution →
15. Q135 CBSE 2026
    1. (i)Why do covalent compounds have low melting and ​boiling points ?
    2. (ii)Name the ​reagent that can ​convert ethanol ​to ethanoic acid.
    3. (iii)Why is ethanoic acid a weaker acid ​than mineral acids ​?
    4. (iv)Define esterification reaction.
    5. (v)What are ​‘Micelles’ ?
    Show answer

    Weak forces between molecules give low melting and boiling points; alkaline ${\text{KMnO}}_4$ oxidises ethanol; ethanoic acid ionises partially; micelles are clusters of soap molecules.

    (i) Within a molecule the atoms are held by strong covalent bonds, but the forces of attraction **between** molecules are weak. Melting and boiling only separate whole molecules from one another, so only these weak intermolecular forces have to be overcome and very little heat energy is needed - hence the low melting and boiling points. (ii) Alkaline potassium permanganate (${\text{KMnO}}_4$ with $\text{KOH}$ or $\text{NaOH}$, warmed); acidified potassium dichromate (${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ with ${\text{H}}_2{\text{SO}}_4$) is equally acceptable. Both are oxidising agents: ${\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{alkaline KMnO}}_4,\text{heat}}{\to }{\text{CH}}_3\text{COOH}$. (iii) Because ethanoic acid ionises only partially in water: ${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$. A mineral acid such as $\text{HCl}$ ionises almost completely, so at the same concentration it gives many more hydrogen ions; fewer ${\text{H}}^{+}$ ions means a weaker acid. (iv) Esterification is the reaction of a carboxylic acid with an alcohol, in the presence of a few drops of concentrated sulphuric acid as catalyst, to give a sweet-smelling ester and water: ${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$. (v) Micelles are the spherical clusters into which soap molecules gather in water, with the hydrophobic hydrocarbon tails pointing inwards towards the oily dirt and the hydrophilic $-{\text{COO}}^{-}$ ends pointing outwards into the water. They hold the dirt suspended as an emulsion so that it can be rinsed away.

    Step-by-step solution →
16. Q136 CBSE 2026
    1. (i)Define the following terms with an example : (I) ​Unsaturated hydrocarbons (II) ​Substitution reaction (III) Saponification
    2. (ii)​Why is ethanoic acid ​called ​glacial ​acetic acid ? Write one use of ethanoic acid.
    Show answer

    Unsaturated hydrocarbons have a double or triple bond; substitution swaps atoms; saponification splits an ester with alkali; pure ethanoic acid freezes at 290 K.

    (i) (I) Unsaturated hydrocarbons are hydrocarbons that contain at least one carbon-carbon double or triple bond, e.g. ethene, ${\text{CH}}_2={\text{CH}}_2$, or ethyne, $\text{HC}\equiv \text{CH}$. (II) A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group, e.g. ${\text{CH}}_4+{\text{Cl}}_2\stackrel{\text{sunlight}}{\to }{\text{CH}}_3\text{Cl}+\text{HCl}$. (III) Saponification is the reaction of an ester with an alkali (sodium hydroxide) to give an alcohol and the sodium salt of the carboxylic acid; with fats or oils the salt formed is soap. E.g. ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$. (ii) Pure ethanoic acid melts at 290 K (about $17^{\circ }\text{C}$), so it often freezes into ice-like crystals during winter in cold climates; hence it is called glacial acetic acid. Use: a 5-8% solution of ethanoic acid in water is vinegar, used as a preservative in pickles.

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17. Q137 CBSE 2025
    Given below ​are the structures of ​some hydrocarbons. ​Select the ​two structures which are ​related ​to each other from the given options :

    1. (a)(i) and (iv)
    2. (b)(ii) and (iv)
    3. (c)(ii) and (iii)
    4. (d)(i) and (iii)
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    (c) (ii) and (iii)

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18. Q138 CBSE 2025
    Select from the ​following a hydrocarbon ​having ​one $\text{C}-\text{C}$ bond and one $\text{C}\equiv \text{C}$ ​bond :

    1. (a)Benzene
    2. (b)Cyclohexane
    3. (c)Butyne
    4. (d)Propyne
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    (d) Propyne

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19. Q139 CBSE 2025
    A Hydrocarbon which does ​not belong ​to the same ​homologous ​series of carbon compounds ​is

    1. (a)${\text{C}}_4{\text{H}}_{10}$
    2. (b)${\text{C}}_6{\text{H}}_{14}$
    3. (c)${\text{C}}_7{\text{H}}_{14}$
    4. (d)${\text{C}}_{10}{\text{H}}_{22}$
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    (c) ${\text{C}}_7{\text{H}}_{14}$

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20. Q140 CBSE 2025
    The ​number of single and ​double bonds ​present ​in a ​molecule ​of benzene (${\text{C}}_6{\text{H}}_6$) respectively, are :

    1. (a)$3$ and $3$
    2. (b)$3$ and $6$
    3. (c)$6$ and $9$
    4. (d)$3$ and $9$
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    (d) $3$ and $9$

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